Question

1. when methane is burned with oxygen the products are carbon dioxide and water. if you produce 36 grams of water and 44 grams of carbon dioxide from 16 grams of methane, how many grams of oxygen were needed for the reaction?

a) 64
b) 32
c) 96
d) 80
e) none of the above

1. which type of energy is associated with motion?

a) kinetic
b) electrical
c) potential
d) chemical
e) none of the above

1. how many joules are there in a 255 calorie snack bar?

a) 2.55 x 10⁵
b) 1.07 x 10⁶
c) 6.09 x 10⁴
d) 1.07x 10³
e) none of the above

1. the boiling point of water is

(1) 212°f (2) 0°c (3) 373 k
a) 1 and 3 only
b) 1 and 2 only
c) 2 and 3 only
d) all of 1, 2, and 3
e) none of 1, 2, and 3

1. what is the value of 27°c on the fahrenheit temperature scale?

a) -6.8
b) 106
c) 81
d) 300
e) none of the above

1. a 15.0 gram lead ball at 25.0°c was heated with 40.5 joules of heat. given the specific heat of lead is 0.128 j/g °c, what is the final temperature of the lead?

a) 21.1°c
b) 46.1°c
c) 0.844°c
d) 77.8°c
e) none of the above

Explanation:

Question 23

Step1: Apply Law of Conservation of Mass

The total mass of reactants (methane + oxygen) equals the total mass of products (water + carbon dioxide). Let the mass of oxygen be \( m_{O_2} \).
\( m_{CH_4} + m_{O_2} = m_{H_2O} + m_{CO_2} \)

Step2: Substitute known values

We know \( m_{CH_4} = 16 \, \text{g} \), \( m_{H_2O} = 36 \, \text{g} \), \( m_{CO_2} = 44 \, \text{g} \).
\( 16 + m_{O_2} = 36 + 44 \)

Step3: Solve for \( m_{O_2} \)

\( 16 + m_{O_2} = 80 \)
\( m_{O_2} = 80 - 16 = 64 \, \text{g} \)

Kinetic energy is defined as the energy associated with motion. Electrical energy is from electric charges, potential energy from position/condition, and chemical energy from chemical bonds. So the energy with motion is kinetic.

Step1: Recall the conversion factor

1 calorie (cal) = 4.184 joules (J).

Step2: Calculate joules from calories

For 255 cal, \( E = 255 \times 4.184 \)
\( E = 255 \times 4.184 \approx 1066.92 \approx 1.07 \times 10^3 \)? Wait, no, 2554.184: 2554=1020, 2550.184=46.92, total 1066.92 ≈ 1.07×10³? Wait, no, 1066.92 is ~1.07×10³? Wait, 1066.92 is 1.06692×10³, which is ~1.07×10³? Wait, no, the options: A)2.55×10⁵, B)1.07×10⁶, C)6.09×10⁴, D)1.07×10³, E)none. Wait, 255 cal 4.184 J/cal = 2554.184 = let's calculate 2554 = 1020, 2550.184 = 255(0.1+0.08+0.004)=25.5 + 20.4 + 1.02=46.92. So total 1020+46.92=1066.92 J ≈ 1.07×10³ J? Wait, no, 1066.92 is 1.06692×10³, which is option D? Wait, no, 1.07×10³ is 1070, which is close. Wait, maybe the conversion is 1 Calorie (food calorie) = 1 kcal = 4184 J. Wait, the question says "calorie" – if it's food calorie (kilocalorie), then 255 kcal 4184 J/kcal = 2554184 = let's calculate 2554000=1,020,000; 255184=46,920; total 1,066,920 J = 1.07×10⁶ J. Ah, that's the mistake! The snack bar's calorie is kilocalorie (food calorie), so 1 cal (food) = 1 kcal = 4184 J. So:

Step1: Correct conversion factor

1 food calorie (cal) = 4184 J.

Step2: Calculate joules

\( E = 255 \times 4184 \)
\( 255 \times 4184 = 255 \times (4000 + 184) = 255×4000 + 255×184 = 1,020,000 + 46,920 = 1,066,920 \, \text{J} \approx 1.07 \times 10^6 \, \text{J} \)

Answer:

A) 64

Question 24