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marta [7]
4 years ago
11

The chemical reaction 2Cu + O2 → 2CuO + 315 kJ is an _____ reaction. Endothermic, combustion Exothermic, combustion Endothermic,

single replacement Exothermic, single replacement
Chemistry
2 answers:
Kobotan [32]4 years ago
8 0

Answer: Exothermic

Explanation: Combustion reaction is a chemical reaction in which hydrocarbon burns in the presence of oxygen to form carbon dioxide, water and energy.

Exothermic reactions are those in which energy is released and endothermic reactions are those in which energy is absorbed.

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution.

2Cu+O_2\rightarrow 2CuO+315kJ

As in the given reaction, the energy is released thus reaction s exothermic.

makvit [3.9K]4 years ago
4 0

the awnser is exothermic (combustion)


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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.)

x \approx 1.19929 \times 10^{-5}.

Hence, the concentration of \rm H^{+} at the equivalence point of this titration would be approximately 1.19929 \times 10^{-5}\; \rm M.

Hence, the pH at the equivalence point of this titration would be:

\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}.

5 0
3 years ago
Help please?? I’m confused
Nady [450]
It’s c. O stands for oxygen and h stands for hydrogen. They both do play in the roll
3 0
3 years ago
Please answer asap MUCH NEEDED!!
gulaghasi [49]

Answer:

I think its C I am sorry if I am wrong

3 0
3 years ago
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