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hoa [83]
3 years ago
11

How much heat does a 100. g sample of copper absorb when its temperature increases by 30.0°C? The specific heat of copper is 0.3

9 J/g °C.
a) 2340 J
b) 1170 J
c) 2.34 J
d) 1.17 J
Chemistry
1 answer:
erica [24]3 years ago
7 0

Answer:

\boxed {\boxed {\sf B. \ 1170 \ Joules }}

Explanation:

We are asked to find how much heat a sample of copper absorbs when the temperature is increased.

Since we know the mass, temperature increase, and specific heat capacity, we can use the following formula to calculate heat.

q= mc \Delta T

The mass of the copper sample is 100 grams, the temperature is changed or increased by 30.0 degrees Celsius, and the specific heat of copper is 0.39 Joules per gram degrees Celsius.

  • m= 100 g
  • c= 0.39 J/g °C
  • ΔT= 30.0 °C

Substitute the values into the formula.

q= (100 \ g )(0.39 \ J/g \textdegree C ) (30.0 \textdegree C )

Multiply the first two values. Note that the units of grams cancel.

q= 39 \ J/ \textdegree C (30.0 \textdegree C )

Multiply again, this time the units of degrees Celsius cancel.

q= 1170 \ J

The copper sample absorbs <u>1170 Joules</u> of heat and <u>Choice B </u>is correct.

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Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

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