Chemistry Problems for 8 – 31 – 16

 

Problem 1

Which particle has no charge?
(1) electron
(2) neutron
(3) positron
(4) proton

¿Qué partícula no tiene cargas?
(1) electrón
(2) neutrón
(3) positrón
(4) protón

Discussion and Solution:

From straight memory:
Electrons have a charge of – 1.
Neutrons have no charge,
Positrons are positively charged electrons. Hence they have a charge of + 1.
Protons have a charge of + 1.
Choice (2) is the correct choice.

Problem 2

Which quantity represents the number of protons in an atom?
(1) atomic number
(2) oxidation number
(3) number of neutrons
(4) number of valence electrons

¿Qué cantidad representa el número de protones en un átomo?
(1) número atómico
(2) número de oxidación
(3) número de neutrones
(4) número de valencia de electrones

Discussion and Solution:
From the definitions of each choice:
Atomic number represents the number of protons in the nucleus of an atom.
Oxidation number represents
Neutrons are neutrally charged particles located in the nucleus. Their quantity has, basically, no bearing on or by the number of protons.
Valence electrons are those electrons located in the outermost electron cloud of an atom.
Choice (1) is the correct choice.

Problem 3

The element sulfur is classified as a
(1) metal
(2) metalloid
(3) nonmetal
(4) noble gas

El elemento azufre es clasificado como un
(1) metal
(2) metaloide
(3) no metal
(4) gas noble

Discussion and Solution:
On the Periodic Table of the Elements,
Metals are located to the left of the “staircase;”
Metalloids are located on the “staircase;”
Nonmetals are located to the right of the “staircase;” and
Noble Gases are located in the rightmost column (Group) on the table.
Choice (3) is the correct choice.

Problem 4

The elements in Group 2 have similar chemical properties because each atom of these elements has the same
(1) atomic number
(2) mass number
(3) number of electron shells
(4) number of valence electrons

Los elementos del Grupo 2 tienen propiedades químicas similares porque cada átomo de esos elementos tiene el mismo
(1) número atómico
(2) número de masa
(3) número de capa de electrones
(4) número de valencia de electrones

Discussion and Solution:
The number of valence electrons (the electrons in the outermost electron cloud) determine, for the most part, the chemical properties of an atom.
Each member of a Group has the same number of valence electrons.
Choice (4) is the correct choice.

Problem 5

Which substance cannot be broken down by a chemical change?
(1) ammonia
(2) arsenic
(3) ethane
(4) propanal

¿Qué sustancia no puede ser quebrada por un cambio químico?
(1) amoníaco
(2) arsénico
(3) etano
(4) propanal

Discussion and Solution:
Compounds may be broken down by chemical changes.
Elements can be broken down only by nuclear changes.
Choices (1), (3) and (4) are compounds.
Choice (2), Arsenic, lists an element.
Choice (2) is the correct choice.

 

Math Problems for 8 – 29 – 16

Problem 1

If f(x) = 3x and g(x) = 2x + 5, at which value of x is f(x) < g(x)?
(1) – 1
(2) 2
(3) – 3
(4) 4

Si f(x) = 3x y g(x) = 2x + 5, ¿en qué valor de x es f(x) < g(x)?
(1) – 1
(2) 2
(3) – 3
(4) 4

 Discussion and Solution:
Try this one without my help. Solve it and send me, via the comment section below, what you determined to be the correct answer.

Problem 2

The function V(t) = 1350(1.017)t represents the value V(t), in dollars, of a comic book t years after its purchase. The yearly rate of appreciation of the comic book is
(1) 17%
(2) 1.7%
(3) 1.017%
(4) 0.017%

La función V(t) = 1350(1.017)t representa el valor V(t), en dólares, de una historieta t años después de su compra. La tasa anual de apreciación de la historieta es
(1) 17%
(2) 1.7%
(3) 1.017%
(4) 0.017%

Discussion and Solution:
Try this one without my help. Solve it and send me, via the comment section below, what you determined to be the correct answer.

Problem 3

A student is asked to solve the equation 4(3x – 1)2 – 17 = 83. The student’s solution to the problem starts as

4(3x – 1)2 = 100

(3x – 1)2 = 25

A correct next step in the solution of the problem is
(1) 3x – 1 = ±5
(2) 3x – 1 = ±25
(3) 9x2 – 1 = 25
(4) 9x2 – 6x + 1= 5

Se le pidió a un estudiante que resolviera la ecuación 4(3x – 1)2 – 17 = 83. La solución dada por el estudiante sobre el problema comienza con

4(3x – 1)2 = 100

(3x – 1)2 = 25

El siguiente paso correcto en la solución del problema es
(1) 3x – 1 = ±5
(2) 3x – 1 = ±25
(3) 9x2 – 1 = 25
(4) 9x2 – 6x + 1= 5

 Discussion and Solution:
Try this one without my help. Solve it and send me, via the comment section below, what you determined to be the correct answer.

 Problem 4

What are the solutions to the equation x2 – 8x = 24?
(1) x = 4 ± 2
(2) x = – 4 ± 2
(3) x = 4 ± 2
(4) x = – 4 ± 2

¿Cuáles son las soluciones a la ecuación x2 – 8x = 24?
(1) x = 4 ± 2
(2) x = – 4 ± 2
(3) x = 4 ± 2
(4) x = – 4 ± 2

 Discussion and Solution:
Try this one without my help. Solve it and send me, via the comment section below, what you determined to be the correct answer.

 Problem 5

Natasha is planning a school celebration and wants to have live music and food for everyone who attends. She has found a band that will charge her $750 and a caterer who will provide snacks and drinks for $2.25 per person. If her goal is to keep the average cost per person between $2.75 and $3.25, how many people, p, must attend?
(1) 225 < p < 325
(2) 325 < p < 750
(3) 500 < p < 1000
(4) 750 < p < 1500

 Natasha está preparando una fiesta escolar y quiere tener una banda en vivo y comida para todos los que asistan. Consiguió una banda que le cobrará $750 y un servicio de banquetes que le proporcionará refrigerios y bebidas a $2.25 por persona. Si su objetivo es mantener el costo promedio por persona entre $2.75 y $3.25, ¿cuántas personas, p, deben asistir?
(1) 225 < p < 325
(2) 325 < p < 750
(3) 500 < p < 1000
(4) 750 < p < 1500

Discussion and Solution:
Try this one without my help. Solve it and send me, via the comment section below, what you determined to be the correct answer.

 

 

Physics Problems for 8 – 26 – 16

Problem 1

As a car approaches a pedestrian crossing the road, the driver blows the horn. Compared to the sound wave emitted by the horn, the sound wave detected by the pedestrian has a
(1) higher frequency and a lower pitch
(2) higher frequency and a higher pitch
(3) lower frequency and a higher pitch
(4) lower frequency and a lower pitch

Mientras un carro se acerca a un peatón cruzando la calle, el conductor toca la corneta. Comparado con la onda de sonido emitida por la corneta, la onda de sonido detectada por el peatón tiene
(1) mayor frecuencia y menor tono
(2) mayor frecuencia y un mayor y tono
(3) menor frecuencia y un mayor tono
(4) menor frecuencia y un menor tono

Discussion and Solution:
Using the concept of the Doppler effect, we find that both the frequency and pitch increase since the car and the pedestrian are moving closer together. Choice (2).

Problem 2

When air is blown across the top of an open water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced. This phenomenon is called
(1) diffraction
(2) refraction
(3) resonance
(4) the Doppler effect

Cuando se sopla aire a través del tope de una botella de agua abierta, las moléculas de aire vibran a una frecuencia particular y se produce sonido. Este fenómeno se llama
(1) difracción
(2) refracción
(3) resonancia
(4) efecto Doppler

Discussion and Solution:
To answer this question, we need to know the definition or concept of each of the terms.
For choice (1), diffraction refers to the bending of sound as the air is moved through a narrow slit.
For choice (2), refraction refers to the bending of waves as they pass from one medium into another. This is usually associated with light waves, although it applies to sound waves also.
For choice (3), resonance refers to the creation of a sound that is caused by another sound (or mechanical movement of an object, e.g., violin strings). Bingo. Choice (3) is correct.
For choice (4), the Doppler effect refers to the change in the frequency of a wave as it moves toward or away from another object.

Problem 3

Which force is responsible for producing a stable nucleus by opposing the electrostatic force of repulsion between protons?
(1) strong
(2) weak
(3) frictional
(4) gravitational

¿Cuál fuerza es responsable de producir un núcleo estable al oponer la fuerza electroestática de repulsión entre protones?
(1) fuerte
(2) débil
(3) friccional
(4) gravitacional

Discussion and Solution:
Since protons all carry a +1 charge, in order for them to remain in close proximity to one another, e.g., in the nucleus of an atom, a strong force is necessary. Of those choices listed, choice (1) is most appropriate.

Problem 4

What is the total energy released when 9.11 x 10-31 kilogram of mass is converted into energy?
(1) 2.73 x 10-22 J
(2) 8.20 x 10-14 J
(3) 9.11 x 10-31 J
(4) 1.01 x 10-47 J

¿Cuál es la energía total liberada cuando 9.11 x 10-31 kilogramos de masa son convertidos en energía?
(1) 2.73 x 10-22 J
(2) 8.20 x 10-14 J
(3) 9.11 x 10-31 J
(4) 1.01 x 10-47 J

Discussion and Solution:
To answer this question, we employ Einstein’s Equation, E = mc2.

E = mc2

E = (9.11 x 10-31 kg)(3.00 x 108 m/s)2

E = 8.90 x 10-14 J

Choice (2).

Problem 5

A shopping cart slows as it moves along a level floor. Which statement describes the energies of the cart?
(1) The kinetic energy increases and the gravitational potential energy remains the same.
(2) The kinetic energy increases and the gravitational potential energy decreases.
(3) The kinetic energy decreases and the gravitational potential energy remains the same.
(4) The kinetic energy decreases and the gravitational potential energy increases.

Un carrito de compras se ralentiza mientras se mueve a lo largo de un piso nivelado. ¿Qué declaración describe la energía del carrito?
(1) La energía cinética aumenta y la energía potencial gravitacional permanece igual.
(2) La energía cinética aumenta y la energía potencial gravitacional disminuye.
(3) La energía cinética disminuye y la energía potencial gravitacional permanece igual.
(4) La energía cinética disminuye y la energía potencial gravitacional aumenta.

Discussion and Solution:
Kinetic energy is a measure of “the energy of motion.” It is determined via the equation:

KE = ½ mv2

Since the cart is slowing down, v must be decreasing. This causes a decrease in KE.

Gravitational potential energy is a measure of “the energy due to position.” It is determined via the equation:

PE = mgh

Since g is a constant, as is the height above the ground, PE must also be constant.
Choice (3).

 

Chemistry Problems for 8 – 24 – 16

Problem 1

Given the equation representing a system at equilibrium:

N2(g) + 3H2(g) ↔ 2NH3(g)

Which statement describes this reaction at equilibrium?
(1) The concentration of N2(g) decreases.
(2) The concentration of N2(g) is constant.
(3) The rate of the reverse reaction decreases.
(4) The rate of the reverse reaction increases.

Dada la ecuación representando un sistema en equilibrio:

N2(g) + 3H2(g) ↔ 2NH3(g)

¿Qué declaración describe esta reacción en equilibrio?
(1) La concentración de N2(g) disminuye.
(2) La concentración de N2(g) es constante.
(3) La velocidad de la reacción inversa disminuye.
(4) La velocidad de la reacción inversa aumenta.

Discussion and Solution:

Since the reaction is at equilibrium, all of the gases, both reactants and products, have a constant concentration. Choice (2).

Problem 2

The acidity or alkalinity of an unknown aqueous solution is indicated by its
(1) pH value
(2) electronegativity value
(3) percent by mass concentration
(4) percent by volume concentration

La acidez o alcalinidad de una solución acuosa desconocido es indicada por su
(1) valor pH
(2) valor de electronegatividad
(3) concentración por porcentual por masa
(4) concentración porcentual por volumen

Discussion and Solution:

pH is defined as the concentration of the Hydrogen ion concentration. Having a pH < 7 is acidic, a pH > 7 is alkaline, and a pH = 0 is neutral. Choice (1).

Problem 3

The laboratory process in which the volume of a solution of known concentration is used to determine the concentration of another solution is called
(1) distillation
(2) fermentation
(3) titration
(4) transmutation

El proceso de laboratorio en el cual el volumen de una solución de concentración conocida es usado para determinar la concentración de otra solución se llama
(1) destilación
(2) fermentación
(3) titulación
(4) transmutación

Discussion and Solution:
In a titration, the formula MaVa = MbVb is used. In this formula, Ma is the molarity of one solution, whereas Va is its volume. Mb is the molarity of the other solution, whereas Vb is its volume. Knowing three of the four variables, along with a little elementary algebra, the fourth variable can easily be determined. Choice (3).

Problem 4

Which list of nuclear emissions is arranged in order from the greatest penetrating power to the least penetrating power?
(1) alpha particle, beta particle, gamma ray
(2) alpha particle, gamma ray, beta particle
(3) gamma ray, alpha particle, beta particle
(4) gamma ray, beta particle, alpha particle

¿Cuál lista de emisiones nucleares está organizada en orden desde el mayor poder de penetración hacia el menor poder de penetración?
(1) partícula alfa, partícula beta, rayos gama
(2) partícula alfa, rayos gama, partícula beta
(3) rayos gama, partícula alfa, partícula beta
(4) rayos gama, partícula beta, partícula alfa

Discussion and Solution:
By straight memorization, for the three nuclear emissions listed, gamma rays have the greatest penetrating power. Alpha particles have the least. Beta particles are somewhere in between. Choice (4).

Problem 5

Which electron shell contains the valence electrons of a radium atom in the ground state?
(1) the sixth shell
(2) the second shell
(3) the seventh shell
(4) the eighteenth shell

¿Cuál capa de electrón contiene las valencias de electrón de un átomo de radio en el estado fundamental??
(1) la sexta capa
(2) la segunda capa
(3) la séptima capa
(4) la decimoctava capa

Discussion and Solution:
From the Periodic Table of the Elements, we can see that Radium (Ra) has an atomic number of 88. Depending upon the completeness of the Periodic Table, we can see that the electron configuration is 2 – 8 – 18 – 32 – 18 – 8 – 2. The outermost shell contains the two valence electrons. Shell 7, Choice (3).

 

Math Problems for 8 – 22 – 16

Problem 1

A satellite television company charges a one-time installation fee and a monthly service charge. The total cost is modeled by the function y = 40 + 90x. Which statement represents the meaning of each part of the function?
(1) y is the total cost, x is the number of months of service, $90 is the installation fee, and $40 is the service charge per month.
(2) y is the total cost, x is the number of months of service, $40 is the installation fee, and $90 is the service charge per month.
(3) x is the total cost, y is the number of months of service, $40 is the installation fee, and $90 is the service charge per month.
(4) x is the total cost, y is the number of months of service, $90 is the installation fee, and $40 is the service charge per month.

Una compañía de televisión satelital cobra una tarifa única de instalación y un cargo por el servicio mensual. El costo total se ejemplifica mediante la función y = 40 + 90x. ¿Qué enunciado representa el significado de cada parte de la función?
(1) y es el costo total, x es la cantidad de meses de servicio, $90 es la tarifa de instalación, y $40 es el cargo por el servicio mensual.
(2) y es el costo total, x es la cantidad de meses de servicio, $40 es la tarifa de instalación, y $90 es el cargo por el servicio mensual.
(3) x es el costo total, y es la cantidad de meses de servicio, $40 es la tarifa de instalación, y $90 es el cargo por el servicio mensual.
(4) x es el costo total, y es la cantidad de meses de servicio, $90 es la tarifa de instalación, y $40 es el cargo por el servicio mensual.

Discussion and Solution:

Since the 40 is listed as being a constant, it is used only a single time, in keeping with the installation charge.

The total value of the last part of the equation is dependent upon the overall value of x. This would be in keeping with the charge for the monthly service.

Charge (2).

Problem 2

If 4x2 – 100 = 0, the roots of the equation are
(1) -25 and 25
(2) -25, only
(3) -5 and 5
(4) -5, only

Si 4x2 – 100 = 0, las raíces de la ecuación son
(1) -25 y 25
(2) -25, solamente
(3) -5 y 5
(4) -5, solamente

Discussion and Solution:

We first add 100 to both sides, getting 4x2 = 100.

We next divide both sides by 4, getting x2 = 25.

Finally, we take the square root of each side, getting x = ±5.

Choice (3).

Problem 3

Which point is not on the graph represented by y = x2 + 3x – 6?
(1) (-6,12)
(2) (-4,-2)
(3) (2,4)
(4) (3,-6)

¿Qué punto no figura en el gráfico representado por y = x2 + 3x – 6?
(1) (-6,12)
(2) (-4,-2)
(3) (2,4)
(4) (3,-6)

Discussion and Solution:

Perhaps the quickest way to solve this dilemma is to try each of the possible sets in the original equation to see which one doesn’t satisfy the equation. Remember, the first number expressed [between the left parenthesis and the comma] is the possible value of x, where the second number [between the comma and the right parenthesis].

For choice (1):

12 = (-6)2 + 3(-6) – 6
12 = 36 – 18 – 6
12 = 12

This one works.

For choice (2):

-2 = (-4)2 + 3(-4) – 6
-2 = 16 – 12 – 6
-2 = -2

This one works.

For choice (3):

4 = (2)2 +3(2) – 6
4 = 4 + 6 – 6
4 = 4

This one works.

For choice (4):

-6 = (3)2 + 3(2) – 6
-6 = 9 + 6 – 6
-6 = 9

Bingo. This set doesn’t work, hence is the correct choice.

Problem 4

A company produces x units of a product per month, where C(x) computations, represents the total cost and R(x) represents the total revenue for the month. The functions are modeled by C(x) = 300x + 250 and R(x) = -0.5x2 + 800x – 100. The profit is the difference between revenue and cost where P(x) = R(x) – C(x). What is the total profit, P(x), for the month?
(1) P(x) = – 0.5x2 + 500x – 150
(2) P(x) = – 0.5x2 + 500x – 350
(3) P(x) = – 0.5x2 – 500x + 350
(4) P(x) = – 0.5x2 + 500x + 350

Una compañía produce x unidades de un producto por mes, para sus cálculos. donde C(x) representa el costo total y R(x) representa los ingresos totales para el mes. Las funciones se ejemplifican mediante C(x) = 300x + 250 y R(x) = – 0.5x2 + 800x – 100. La ganancia es la diferencia entre los ingresos y el costo donde P(x) = R(x) – C(x). ¿Cuál es la ganancia total, P(x), para el mes?
(1) P(x) = – 0.5x2 + 500x – 150
(2) P(x) = – 0.5x2 + 500x – 350
(3) P(x) = – 0.5x2 – 500x + 350
(4) P(x) = – 0.5x2 + 500x + 350

Discussion and Solution:

We start off with the combined equation, P(x) = R(x) – C(x), and substitute in the values for R(x) and C(x) and, finally, solve for P(x).

P(x) = R(x) – C(x)
P(x) = [- 0.5x2 + 800x – 100] – [300x + 250]
P(x) = – 5x2 + 800x – 100 – 300x – 250
P(x) = -5x2 + 500x – 350

Choice (2).

Problem 5

Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy’s age, j, if he is the younger man?
(1) j2 + 2 = 783
(2) j2 – 2 = 783
(3) j2 + 2j = 783
(4) j2 – 2j = 783

Las edades de Sam y Jeremy son números enteros impares consecutivos. El producto de sus edades es 783. ¿Qué ecuación podría usarse para encontrar la edad de Jeremy, j, si él es el más joven?
(1) j2 + 2 = 783
(2) j2 – 2 = 783
(3) j2 + 2j = 783
(4) j2 – 2j = 783

Discussion and Solution:

From the first sentence, we get:

s = j + 2.

From the second sentence, we get:

783 = sj.

By combining the two equations, we get:

783 = sj
783 = (j + 2)j
783 = j2 + 2j

Choice (3).

 

Physics Problems for 8 – 19 – 16

Problem 1

A 2.0-kilogram mass is located 3.0 meters above the surface of Earth. What is the magnitude of Earth’s gravitational field strength at this location?
(1) 4.9 N/kg
(2) 2.0 N/kg
(3) 9.8 N/kg
(4) 20. N/kg

Una masa de 2.0-kilogramos está localizada 3.0 metros por encima de la superficie de la Tierra. ¿Cuál es la magnitud de la fortaleza del campo gravitacional de la Tierra en esta ubicación?
(1) 4.9 N/kg
(2) 2.0 N/kg
(3) 9.8 N/kg
(4) 20. N/kg

Discussion and Solution:

Earth’s gravitational field strength is a constant, 9.8 N/kg. Choice (3)

Problem 2

A truck, initially traveling at a speed of 22 meters per second, increases speed at a constant rate of 2.4 meters per second2 for 3.2 seconds. What is the total distance traveled by the truck during this 3.2-second time interval?
(1) 12 m
(2) 58 m
(3) 70. m
(4) 83 m

Un camión, viajando inicialmente a una velocidad de 22 metros por segundo, aumenta velocidad a un ritmo constante de 2.4 metros por segundo2 por 3.2 segundos. ¿Cuál es la distancia total viajada por el camión durante este intervalo de tiempo de 3.2 segundos?
(1) 12 m
(2) 58 m
(3) 70. m
(4) 83 m

Discussion and Solution:

d = vit + ½at2

d = (22 m/s)(3.2 s) + ½(2.4 m/s2)(3.2 s)2

d = 70.4 m + 12.288 m

d = 82.688 m

 

This rounds to 83 m (two significant figures). Choice (4).

 

Problem 3

A 750-newton person stands in an elevator that is accelerating downward. The upward force of the elevator floor on the person must be
(1) equal to 0 N
(2) less than 750 N
(3) equal to 750 N
(4) greater than 750 N

Una persona de 750-newton se para en un elevador que acelera hacia abajo. La fuerza del elevador hacia arriba en la persona debe ser
(1) igual a 0 N
(2) menor que 750 N
(3) igual a 750 N
(4) mayor que 750 N

Discussion and Solution:

Since the elevator is accelerating downward, the upward force of the elevator floor on the person must be less than 750 N. Choice (2).

Problem 4

A 3.0-kilogram object is acted upon by an impulse having a magnitude of 15 newton•seconds. What is the magnitude of the object’s change in momentum due to this impulse?
(1) 5.0 kg•m/s
(2) 15 kg•m/s
(3) 3.0 kg•m/s
(4) 45 kg•m/s

Un objeto de 3.0-kilogramos actúa bajo un impulso teniendo una magnitud de 15 newton•segundos. ¿Cuál es la magnitud del cambio del objeto en el momento debido a este impulso?
(1) 5.0 kg•m/s
(2) 15 kg•m/s
(3) 3.0 kg•m/s
(4) 45 kg•m/s

Discussion and Solution:

J = Fnet t = Δp

 Since Impulse (J) is equal to change in momentum (Δp), the correct answer is Choice (2).

 

Problem 5

An air bag is used to safely decrease the momentum of a driver in a car accident. The air bag reduces the magnitude of the force acting on the driver by
(1) increasing the length of time the force acts on the driver
(2) decreasing the distance over which the force acts on the driver
(3) increasing the rate of acceleration of the driver
(4) decreasing the mass of the driver

Una bolsa de aire es usada para disminuir sin peligro el momento de un conductor en un accidente automovilístico. La bolsa de aire reduce la magnitud de la fuerza que actúa en el conductor al
(1) aumentar la duración del tiempo que la fuerza ejerce en el conductor
(2) disminuir la distancia sobre la cual la fuerza actúa en el conductor
(3) aumentar el ritmo de aceleración del conductor
(4) disminuir la masa del conductor

Discussion and Solution:

p = mv

Since momentum (p) is calculated by the above equation, where m = mass and v = velocity, since the mass of the driver doesn’t change, it’s velocity must decrease in order for the product of the two to decrease.

v = d / t

Since the distance (d) doesn’t change, time (t) must increase in order for the quotient to decrease. Choice (1).

 

Chemistry Problems for 8 – 17 – 16

Problem 1

A neutron has a charge of
(1) +1
(2) +2
(3) 0
(4) − 1

Un neutrón tiene una carga de
(1) +1
(2) +2
(3) 0
(4) −1

Discussion and Solution:

Memorization. Neutrons are neutral. Choice (3).

Problem 2

Which particle has the least mass?
(1) alpha particle
(2) beta particle
(3) neutron
(4) proton

¿Cuál partícula tiene la menor masa?
(1) partícula alfa
(2) partícula beta
(3) neutrón
(4) protón

Discussion and Solution:

Memorization. An alpha particle has a mass of 2. A beta particle 0. A neutron 1. A proton 1. Choice (2).

Problem 3

A sample of matter must be copper if
(1) each atom in the sample has 29 protons
(2) atoms in the sample react with oxygen
(3) the sample melts at 1768 K
(4) the sample can conduct electricity

Una muestra de material debe ser cobre si
(1) cada átomo en la muestra tiene 29 protones
(2) los átomos en la muestra reaccionan con oxígeno
(3) la muestra se derrite a 1768 K
(4)la muestr a puede conducir electricidad

Discussion and Solution:

According to the Periodic Table of the Elements, copper has an atomic number of 29. This means that copper atoms contain 29 protons.

Problem 4

In the electron cloud model of the atom, an orbital is defined as the most probable
(1) charge of an electron
(2) conductivity of an electron
(3) location of an electron
(4) mass of an electron

En el modelo nube de electrones del átomo, una orbital se define como la más probable
(1) carga de un electrón
(2) conductividad de un electrón
(3) ubicación de un electrón
(4) masa de un electron

Discussion and Solution:

Contrary to atomic models of yesteryear, today we describe the location of an electron as being, statistically, a cloud.

Problem 5

The elements on the Periodic Table are arranged in order of increasing
(1) atomic number
(2) mass number
(3) number of isotopes
(4) number of moles

Los elementos en la Tabla Periódica están ordenados en orden creciente de
(1) número atómico
(2) número de masa
(3) número de isotopos
(4) número de moles

Discussion and Solution:

As one progresses from left to right within a period, or from top to down within a group, atomic numbers increase. Choice (1).

 

Math Problems for 8 – 15 – 16

Problem 1

For which function defined by a polynomial are the zeros of the polynomial −4 and −6?

1)  y = x2 − 10x − 24

2)  y = x2 + 10x + 24

3)  y = x2 + 10x − 24

4)  y = x2 − 10x + 24

Discussion and Solution:

There are two ways of solving this problem – (1) solve each equation for the value of x and (2) substitute each possible solution into each equation to which one works. Let’s try them both.

Method (1) –

  • y = x2 – 10x – 24

0 = x2 – 10x – 24

0 = (x – 4)(x – 6)

0 = (x – 4); x = 4

0 = (x – 6); x = 6

No way

  • y = x2 + 10x + 24

0 = x2 + 10x + 24

0 = (x + 4)(x + 6)

0 = (x + 4); x = – 4

0 = (x + 6); x = – 6

Bingo – they work here – we can stop here since we have the answer

Method (2) –

  • y = x2 – 10x – 24

0 = x2 – 10x – 24

0 = (– 4)2 – 10(– 4) – 24

0 = 16 + 40 – 24

0 ≠ 32

No way

  • y = x2 + 10x + 24

0 = x2 + 10x + 24

0 = (– 4)2 + 10(– 4) + 24

0 = 16 – 40 + 24

0 = 0

This one works, but we better check the other one.

y = x2 + 10x + 24

0 = x2 + 10x + 24

0 = (– 6)2 + 10(– 6) + 24

0 = 36 – 60 + 24

Bingo. This one works also.

Correct solution is (2).

Problem 2

If f(x) = x2 − 2x − 8 and g(x) = ¼ x − 1, for which values of x is f(x) = g(x)?

1) −1.75 and −1.438

2) −1.75 and 4

3) −1.438 and 0

4) 4 and 0

Discussion and Solution:

The easiest way to solve this is to substitute each possible solution into each equation and see which one clicks.

Since options (3) and (4) both have 0 as a possible solution, let’s start with this one and see what happens.

f(x) = x2 − 2x – 8 = (0)2 – 2(0) – 8 = – 8

g(x) = ¼ x – 1 = ¼ (0) – 1 = – 1

I don’t know about you, but I was taught – 8 ≠ – 1. Therefore, choices (3) and (4) are out.

Now let’s try options (1) and (2). Let’s try the 4.

f(x) = x2 − 2x – 8 = (4)2 – 2(4) – 8 = 16 – 8 – 8 = 0

g(x) = ¼ x – 1 = ¼ (4) – 1 = 1 – 1 = 0

 That works, now let’s try the – 1.75.

f(x) = x2 − 2x – 8 = (– 1.75)2 – 2(– 1.75) – 8 = 3.0625 + 3.5 – 8 = – 1.4375

g(x) = ¼ x – 1 = ¼ (– 1.75) – 1 = – 0.4375 – 1 = – 1.4375

That’s okay, but let’s try the – 1.438 to rule out choice (1).

f(x) = x2 − 2x – 8 = (– 1.438)2 – 2(– 1.438) – 8 = 2.0678 + 2.876 – 8 = – 3.056

g(x) = ¼ x – 1 = ¼ (– 1.438) – 1 = – 0.3595 – 1 = – 1.3595

Nope.

Choice (2) is the correct choice.

Problem 3

Which system of equations has the same solution as the system below?

2x + 2y = 16

3x y = 4

1)  2x + 2y = 16

6x − 2y = 4

2)  2x + 2y = 16

6x − 2y = 8

3)  x + y = 16

3x y = 4

4)  6x + 6y = 48

6x + 2y = 8

 Discussion and Solution:

In order to have the same solution, when we divide the possible solutions by some small number, say a 2 or 3, we should arrive at the original equation. Here goes.

Choice (1):

1)  2x + 2y = 16

6x − 2y = 4

The first equation is the same as one of the original equations. As for the second equation, the left side appears to be doubled from the original equation, but not the right side. This choice is out.

Choice (2):

2)  2x + 2y = 16

6x − 2y = 8

The first equation is the same as one of the original equations. As for the second equation, both sides have been doubled as compared to the other original equation. This choice is the correct one.

Choice (3):

3)  x + y = 16

3x y = 4

In this choice, the second equation is the same as one of the original equations, but the first equation is messed up. The right side of this equation is the same as the right side of the other original equation, but the left side has been halved. This choice is out.

Choice (4):

4)  6x + 6y = 48

6x + 2y = 8

In this choice, the first equation is the triple of one of the original equations, but the second one is messed up. At first glance, it appears that the equation has merely been doubled. Take a closer look. If it were just doubled, it would be 6x – 2y = 8, not 6x + 2y = 8. This choice is out.

Choice (2) is the correct one.

Problem 4

Mo’s farm stand sold a total of 165 pounds of apples and peaches. She sold apples for $1.75 per pound and peaches for $2.50 per pound. If she made $337.50, how many pounds of peaches did she sell?

1)  11

2)  18

3)  65

4)  100

 Discussion and Solution:

Let’s set up two equations and solve. a = number pounds of apples. p = number of pounds of peaches.

a + p = 165

1.75a + 2.50p = 337.50

Two equations with two unknowns. Don’t you just love that? As you’re probably aware, when we have two equations with two unknowns, there are three ways of solving this. They are:

(a) Solve one equation for one of the unknowns and substitute this into the other equation and solve.

(b) Graph the two equations and look to see where they intersect. This point of intersection will indicate the solutions. If the lines are parallel or on top of each other, then there is no common solution.

(c) Multiply one of the equations by some number (in this case, multiply the top equation by 1.75) and subtract the two equations, thereby eliminating one of the variables. Solve for the remaining variable. Now, substitute this value into either of the original equations and determine the missing value.

Let’s use procedure (a).

a + p = 165

a = 165 – p

 

1.75a + 2.50p = 337.50

1.75(165 – p) + 2.50p = 337.50

288.75 – 1.75p + 2.50p = 337.50

288.75 + 0.75p = 337.50

0.75p = 337.50 – 288.75 = 48.75

p = 65

Choice (3) is the correct choice.

Problem 5

If A = 3x2 + 5x − 6 and B = −2x2 − 6x + 7, then A B equals

1) −5x2 − 11x + 13

2) 5x2 + 11x − 13

3) −5x2x + 1

4) 5x2x + 1

 Discussion and Solution:

Since we’re looking for AB, we merely need to set the two equations equal to each other.

AB

(3x2 + 5x – 6) – (−2x2 − 6x + 7)

3x2 + 5x – 6 + 2x2 − 6x – 7

5x2 + 11x – 13

Choice (2) is correct.

Physics Problems for 8 – 12 – 16

Problem 1

Which quantity is scalar?
(1) mass
(2) force
(3) momentum
(4) acceleration

¿Qué cantidad es escalar?

(1) masa
(2) fuerza
(3) momento
(4) aceleración

Discussion and Solution

A scalar quantity, as opposed to a vector quantity, has only a magnitude (numerical) component, but no direction component. Choice (1)

Problem 2

What is the final speed of an object that starts from rest and accelerates uniformly at 4.0 meters per second2 over a distance of 8.0 meters?
(1) 8.0 m/s
(2) 16 m/s
(3) 32 m/s
(4) 64 m/s

¿Cuál es la velocidad final de un objeto que comienza desde descanso y acelera uniformemente a 4.0 metros por segundo2 sobre una distancia de 8.0 metros?
(1) 8.0 m/s
(2) 16 m/s
(3) 32 m/s
(4) 64 m/s

Discussion and Solution

Vf2 = Vi2 + 2AD, where Vf = final velocity, Vi = initial velocity, A = acceleration, and D = distance.
Vf2 = Vi2 + 2AD = (0.0 m/s)2 + 2(4.0 m/s2)(8.0 m) = 64 m2/s2
Taking the square root of both sides, we get Vf = 8.0 m/s, Choice (1).

Problem 3

The components of a 15-meters-per-second velocity at an angle of 60.° above the horizontal are
(1) 7.5 m/s vertical and 13 m/s horizontal
(2) 13 m/s vertical and 7.5 m/s horizontal
(3) 6.0 m/s vertical and 9.0 m/s horizontal
(4) 9.0 m/s vertical and 6.0 m/s horizontal

Los componentes de una velocidad de 15 metros por segundo a un ángulo de 60° sobre la horizontal son
(1) 7.5 m/s vertical y 13 m/s horizontal

(2) 13 m/s vertical y 7.5 m/s horizontal

(3) 6.0 m/s vertical y 9.0 m/s horizontal

(4) 9.0 m/s vertical y 6.0 m/s horizontal

Discussion and Solution

Think of a Right Triangle. Think of the definition of Sin and Cos. Sin = Opposite / Hypotenuse. Cos = Adjacent / Hypotenuse.
Sin 60o = X / 15 m/s. X = (15 m/s)Sin60o = 13 m/s vertical.
Cos 60o = X / 15 m/s. X = (15 m/s)Cos 60o = 7.5 m/s horizontal
Choice (2).

Problem 4

What is the time required for an object starting from rest to fall freely 500. meters near Earth’s surface?
(1) 51.0 s
(2) 25.5 s
(3) 10.1 s
(4) 7.14 s

¿Cuál es el tiempo requerido para que un objeto que comienza desde descanso caiga libremente 500 metros cerca de la superficie de la Tierra?
(1) 51.0 s
(2) 25.5 s
(3) 10.1 s
(4) 7.14 s

Discussion and Solution

D = ViT + ½AT2
Since we are starting from rest, Vi = 0.0 m/s. This alters our equation to: D = ½AT2. Solving for T, we get T = SQRT(2D/A) = SQRT(2[500. m] / 9.8 m/s2) = 10.1 s. Choice (3).

Problem 5

A baseball bat exerts a force of magnitude F on a ball. If the mass of the bat is three times the mass of the ball, the magnitude of the force of the ball on the bat is
(1) F
(2) 2F
(3) 3F
(4) F/3

Un bate de béisbol ejerce una fuerza de magnitud F en una bola. Si la masa del bate es tres veces la masa de la bola, la magnitud de la fuerza de la bola en el bate es
(1) F
(2) 2F
(3) 3F
(4) F/3

Discussion and Solution

Remember, a Force exerted on an object one way, is counter balanced by a Force of equal magnitude exerted in the opposite direction. Choice (1).

Chemistry Problems for 8 – 10 – 16

 

Problem 1

Which particles have approximately the same mass?

(1)       alpha particle and beta particle

(2)       alpha particle and proton

(3)       neutron and positron

(4)       neutron and proton

 

¿Cuáles partículas tienen aproximadamente la misma masa?

(1)        La partícula alfa y la partícula beta

(2)        La partícula alfa y protón

(3)        neutrón y positrón

(4)        neutrón y protón

 

Solution and Discussion

To answer this question depends upon your remembering the masses of the various nuclear particles:

Alpha particle = 4

Beta particle = 0

Neutron = 1

Positron = 0

Proton = 1

From the above, a neutron and a proton both have a mass of 1, therefor Choice (4) is the correct choice.

 

Problem 2

Which phrase describes an atom?

(1)       a negatively charged nucleus surrounded by positively charged protons

(2)       a negatively charged nucleus surrounded by positively charged electrons

(3)       a positively charged nucleus surrounded by negatively charged protons

(4)       a positively charged nucleus surrounded by negatively charged electrons

 

¿Qué frase describe un átomo?

(1)        un núcleo cargado negativamente rodeado por protones cargados positivamente

(2)        un núcleo cargado negativamente rodeado por electrones cargados positivamente

(3)        un núcleo cargado positivamente rodeado por protones cargados negativamente

(4)        un núcleo cargado positivamente rodeado por electrones cargados negativamente

 

Solution and Discussion

To answer this problem, you need to remember what an atom consists of.

The nucleus has a positive charge due to the presence of positively charged protons.

The nucleus is surrounded by electrons. Electrons are negatively charged.

Therefor Choice (4) is the correct choice.

 

Problem 3

An orbital is defined as a region of the most probable location of

(1)       an electron

(2)       a neutron

(3)       nucleus

(4)       proton

 

Un orbital se define como una región de la más probable localización de

(1)        un electrón

(2)        un neutrón

(3)        un núcleo

(4)        un protón

 

Solution and Discussion

To answer this problem, you need to remember the definition of an orbital.

An orbital is defined as a region of the most probable location of electrons.

Choice (1) is the correct choice.

 

Problem 4

The bright-line spectrum of an element in the gaseous phase is produced as

(1)       protons move from lower energy states to higher energy states

(2)       protons move from higher energy states to lower energy states

(3)       electrons move from lower energy states to higher energy states

(4)       electrons move from higher energy states to lower energy states

 

El espectro de líneas brillantes de un elemento en la fase gaseosa se produce cuando

(1)        los protones pasan de estados de menor energía a estados de mayor energía

(2)        los protones pasan de estados de mayor energía a estados de energía más bajos

(3)        los electrones se mueven a partir de estados de menor energía a estados de mayor energía

(4)        los electrones se mueven a partir de los estados de mayor energía a estados de energía más bajos

 

Solution and Discussion

The answer to this problem is dependent upon knowing what produces bright-line spectrum.

The bright-line spectrum of an element in the gaseous phase is produced as electrons move from higher energy states to lower energy states. You probably did a lab to show this.

Choice (4) is the correct choice.

 

Problem 5

An atom of lithium-7 has an equal number of

(1)       electrons and neutrons

(2)       electrons and protons

(3)       positrons and neutrons

(4)       positrons and protons

 

Un átomo de litio-7 tiene un número igual de

(1)        electrones y neutrones

(2)        electrones y protones

(3)        positrones y neutrones

(4)        positrones y protones

 

Solution and Discussion

All atoms, being electrically neutral, contain an equal number of positive charges (protons) and negative charges (electrons), Choice (2).