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3 years ago

Intensive or Extensive?

Chemistry

1 answer:

Daniel [21]3 years ago

4 0

Ou this is complicated

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What is the correct name for N2O

krek1111 [17]

Explanation:

Nitrous oxide

^_^^.^^.^

6 0

3 years ago

Read 2 more answers

At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction

Brums [2.3K]

Answer : The equilibrium concentrations of all species $NO,Cl_2\text{ and }NOCl$ are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of $NO$ = 2 mole

Moles of $Cl_2$ = 1 mole

Volume of solution = 1 L

Initial concentration of $NO$ = 2 M

Initial concentration of $Cl_2$ = 1 M

The given balanced equilibrium reaction is,

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Initial conc. 2 M 1 M 0

At eqm. conc. (2-2x) M (1-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}$

The $K_c$ for reverse reaction = $\frac{1}{1.6\times 10^{-5}}$

Now put all the given values in this expression, we get :

$\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}$

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of $NO,Cl_2\text{ and }NOCl$ at equilibrium are :

Concentration of $NO$ = (2-2x) M = (2 - 2 × 0.975) M = 0.05 M

Concentration of $Cl_2$ = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of $NOCl$ = x M = 0.975 M

Therefore, the equilibrium concentrations of all species $NO,Cl_2\text{ and }NOCl$ are, 0.05 M, 0.043 M and 0.975 M respectively.

6 0

3 years ago

A child sits on top of a slide at the park. He pushes forward lightly, but does not go down the slide. Pick the sentence that

Vinvika [58]

Answer:

The net force on the child is zero

4 0

3 years ago

A buffer solution is prepared by adding NaC2H202

olganol [36]

Shift to reactants(left)

<h3>Further explanation </h3>

A buffer solution is a solution that can maintain a good pH value due to the addition of a little acid or a little base or dilution.

The buffer solution can be acidic or basic

Acid buffer solutions consist of weak acids and their salts.

A buffer solution of NaC2H202 and HC2H2O2 (acetic acid) is included in the acid buffer

So :

a slight addition of acid (H⁺) will be balanced by the conjugate base

the addition of a small base (OH⁻) will be balanced by the weak acid

With the addition of acid (H +), the equilibrium will shift to the left, in the formation of CH3COOH. The added acid will be neutralized by the conjugate base component (CH3COO−).

7 0

3 years ago

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

Sergio [31]

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.

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

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

7. Calculate the pH

$\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}$

4 0

3 years ago