All categories

3 years ago

A 135 g sample of a metal requires 2.50 kJ to change its temperature from 19.5°C to 95.5°C to 100.0°C?

Chemistry

1 answer:

just olya [345]3 years ago

8 0

Answer:

0.23J/g°C

Explanation:

Given parameters:

Mass of sample = 135g

Amount of heat = 2.5kJ

Initial temperature = 19.5°c

Final temperature = 100°C

Unknown:

Specific heat capacity of the metal = ?

Solution:

The specific heat capacity of a substance is the amount of heat required to the raise the temperature of 1g of the substance by 1°C.

H = m C (T₂ - T₁ )

H is the amount of heat

m is the mass

C is the specific heat capacity

T₂ is the final temperature

T₁ is the initial temperature;

2.5 x 10³ = 135 x C x (100 - 19.5)

2500 = 10867.5C

C = $\frac{2500}{10867.5}$ = 0.23J/g°C

You might be interested in

I need help, do you mind helping me?

Sever21 [200]

The mole ratio is 2:19 or 1:9.5

5 0

3 years ago

4. An atom is neutral because it has the same number of protons as it has:

stealth61 [152]

Neutrons the way this question is setup the atom would not be able to be neutral

7 0

3 years ago

Which of the following may suggest a catalyst has been used in a reaction, given the energy diagram for the same reaction withou

anzhelika [568]

The answer I think would be C

5 0

3 years ago

A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equival

stealth61 [152]

Answer:

Approximately $4.92$ .

Explanation:

Initial volume of the solution: $V = 190.0\; \rm mL = 0.1900\; \rm L$ .

Initial quantity of $\rm NH_3$ :

$\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}$ .

Ammonia $\rm NH_3$ reacts with hydrochloric $\rm HCl$ acid at a one-to-one ratio:

$\rm NH_3 + HCl \to NH_4 Cl$ .

Hence, approximately $n({\rm HCl}) = 0.154375\; \rm mol$ of $\rm HCl\!$ molecules would be required to exactly react with the $\rm NH_3\!$ in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that $0.3733\; \rm mol \cdot L^{-1}$ $\rm HCl$ solution required for reaching the equivalence point of this titration:

$\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}$ .

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately $0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L$ .

If no hydrolysis took place, $0.154375\; \rm mol$ of $\rm NH_4 Cl$ would be produced. Because $\rm NH_4 Cl\!$ is a soluble salt, the solution would contain $0.154375\; \rm mol\!$ of $\rm {NH_4}^{+}$ ions. The concentration of $\rm {NH_4}^{+}\!$ would be approximately:

$\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}$ .

However, because $\rm NH_3 \cdot H_2O$ is a weak base, its conjugate $\rm {NH_4}^{+}$ would be a weak base.

$\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}$ .

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

$\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}$ .

Let $x\; \rm mol \cdot L^{-1}$ be the increase in the concentration of $\rm H^{+}$ in this solution because of this reversible reaction. (Notice that $x \ge 0$ .) Construct the following $\text{RICE}$ table:

$\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}$ .

Thus, at equilibrium:

Concentration of the weak acid: $[{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M$ .
Concentration of the conjugate of the weak acid: $[{\rm NH_3}] = x\; \rm M$ .
Concentration of $\rm H^{+}$ : $[{\rm {H}^{+}}] \approx x\; \rm M$ .

$\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}$ .

$\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}$

Solve for $x$ . (Notice that the value of $x\!$ is likely to be much smaller than $0.255782$ . Hence, the denominator on the left-hand side $(0.255782 - x) \approx 0.255782$ .)