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Molodets [167]
2 years ago
14

Which of the following would not cause an increase in the pressure of a gaseous system in a container? A The container is made l

arger. Additional gas is added to the container. B A C The temperature in the container is increased. The volume of the container is decreased.​
Chemistry
2 answers:
ludmilkaskok [199]2 years ago
5 0

Answer:

The container is made larger

Explanation:

Delicious77 [7]2 years ago
5 0

Answer: The volume of the container is increased

Explanation:

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Consider the reaction pathway graph below.
Alex17521 [72]
Hello!

The reaction that the graph represents is A. Exothermic because Hrxn=-167 kJ

To calculate Hrxn we apply the following equation:

Hrxn=Hproducts-Hreagents=-625kJ-(-458kJ)=-167kJ

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction releases energy in the form of heat and that the reaction is exothermic.

Have a nice day!
6 0
3 years ago
Un tanque de acetileno para una antorcha de soldadura de oxiacetileno proporciona 9340 L de gas acetileno, C2H2, a 0°C y 1 atm 2
Helen [10]

Answer:

3.33 tanques de O₂

Explanation:

Basados en la reacción:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>

<em />

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>

Si un tanque contiene 7x10³ L de O₂ serán necesarios:

23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>

6 0
3 years ago
In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
Mamont248 [21]

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

5 0
3 years ago
I need help!! ASAP please..
Marrrta [24]

Answer:

164 g

Explanation:

8 0
2 years ago
Please answer this. Look at the picture for the questions, Thanks!
kenny6666 [7]
The water molecules are moving so slow that they end up sticking together to form a solid
4 0
2 years ago
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