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eimsori [14]
3 years ago
15

Consider the generic reaction: 2 A(g) + B(g) → 2 C(g). If a flask initially contains 1.0 atm of A and 1.0 atm of B, what is the

pressure in the flask if the reaction proceeds to completion? (Assume constant volume and temperature.)a. 1.0 atmb. 1.5 atmc. 2.0 atmd. 3.0 atm
Chemistry
1 answer:
irina1246 [14]3 years ago
3 0

Answer:

b. 1.5 atm.

Explanation:

Hello!

In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:

P=P_A+P_B+P_C

Now, if we introduce the stoichiometry, and the change in the pressure x we can write:

P=1.0-2x+1.0-x+2x

Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:

x=\frac{1.0atm}{2}=0.5atm

Thus, the final pressure is:

P=1.0-2(0.5)+1.0-(0.5)+2(0.5)\\\\P=1.5atm

Therefore the answer is b. 1.5 atm.

Best regards!

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Explanation:

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When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

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When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

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Atypicalaspirintabletcontains325mgofacetylsalicylic acid (HC9H7O4). Calculate the pH of a solution that is prepared by dissolvin
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Answer:

\boxed{2.65}

Explanation:

1. Mass of acetylsalicylic acid (ASA)

m = \text{2 tablets} \times \dfrac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}

2. Moles of ASA

HC₉H₇O₄ =180.16 g/mol

n = \text{750 mg} \times \dfrac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}

3. Concentration of ASA

c = \dfrac{\text{4.163 mmol}}{\text{237 mL}} = \text{0.01757 mol/L}

4. Set up an ICE table

\begin{array}{ccccccc}\text{HA} & + & \text{H$_{2}$O}& \, \rightleftharpoons \, &\text{H$_{3}$O$^{+}$} & + &\text{A}^{-}\\0.01757 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.01757-x & & & &x & & x \\\end{array}\\

5. Solve for x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0

6. Solve the quadratic equation.

a = 1; b = 3.33 \times 10^{-4}; c = -5.851 \times 10^{-6}

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258

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