sophomore chemistry final pt03

Calculate A 50g sample is heated with 500 joules of energy and has a 10*C temperature change. Calculate it’s specific heat.

• 1 j/g*C

• 250000 j/g*C

• 500 j/g*C

• 10 j/g*C

Tags

NGSS.HS-PS3-1

A 24.0 gram sample of copper was heated from 25.0^oC to 500.0^oC, 4378 J of heat were absorbed, what is the specific heat of copper?

• 0.384 J/g°C

• 49909200 J/g°C

• 2.60 J/g°C

• 8.77 J/g°C

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-2

 3,478 kJ into cal

• 832,000 J

• 8,321 kcal

• 3,478,000 J

• 832, 100 cal

Convert 120 Cal to Joules

• .502 Joules
• 502080 Joules
• 502 Joules

• 120 kcal

• 83 kcal

• 315 kcal

• 284 kcal

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

How many kilocalories are required to raise the temper ature of a 35.0 g sample of iron from 25 °C to 35 °C? (c =0.108 cal/g °C)

• 0.0378 kcal

• 3.78 kcal

• 378 kca

• 37.8 kcal

• 8.1 kJ

• 13.5 kJ

• 26.30 kJ

• 54.20 kJ

Tags

NGSS.HS-PS3-1

how many moles of carbon dioxide would be produced from the following reaction if 175 kJ of energy is given off? 2CO + O2 -> 2CO2 ΔH = -566

• 15 moles

• 0.155 moles

• 155 moles

• 15.5 moles

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation:

1/2N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ

1/2N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ

• 22 kJ

• -156.6 kJ

• 15.6 kJ

• -1.55 kJ

Answer explanation

To calculate the enthalpy of the reaction:

2NO(g) + O2(g) -> 2NO2(g)

we can use Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions involved.

First, we need to reverse the first reaction:

NO2(g) -> 1/2N2(g) + O2(g)

Since we want to reverse the reaction, we also need to reverse the sign of the enthalpy change, so:

ΔH°A = -33.2 kJ

Next, we multiply the first reaction by 2 to obtain the desired stoichiometry:

2(NO2(g)) -> N2(g) + 2O2(g)

Since we multiplied the reaction by 2, we also need to multiply the enthalpy change by 2:

2(ΔH°A) = 2(-33.2 kJ) = -66.4 kJ

Next, we reverse the second reaction:

NO(g) -> 1/2N2(g) + 1/2O2(g)

Since we want to reverse the reaction, we also need to reverse the sign of the enthalpy change, so:

ΔH°B = -90.2 kJ

Now, we can write the overall reaction as the sum of these two reactions:

2(NO(g)) + O2(g) -> 2(NO2(g))

The overall enthalpy change of the reaction is the sum of the enthalpy changes of the individual reactions:

ΔH°overall = 2(ΔH°A) + ΔH°B

ΔH°overall = -66.4 kJ + (-90.2 kJ)

ΔH°overall = -156.6 kJ

Therefore, the enthalpy of the reaction 2NO(g) + O2(g) -> 2NO2(g) is -156.6 kJ.

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

Which is smaller a calorie or a joule?

• calorie

• joule