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Aloiza [94]
3 years ago
8

1. The mass of 6.45 x 1022 atoms C​

Chemistry
1 answer:
Fed [463]3 years ago
7 0

Answer:

\large \boxed{\text{1.29 g}}

Explanation:

We must convert atoms of C to moles of C and then to grams of C.

1. Convert atoms to moles

\text{Moles of C} = 6.45 \times10^{22}\text{ atoms C} \times \dfrac{\text{1 mol C}}{6.022 \times\ 10^{23} \text{ atoms C}} = \text{0.1071 mol C}

2. Convert moles to grams

\text{Mass of C} = \text{0.1071 mol C} \times \dfrac{\text{12.01 g C}}{\text{1 mol C}} = \text{1.29 g C}\\\text{The mass of C is $\large \boxed{\textbf{1.29 g}}$}

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3 years ago
An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc
pantera1 [17]

Answer:

5.35 *10^{-4}M

Explanation:

Equation for the reaction is as follows:

2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

By Applying the ICE Table; we have

                             2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

Initial                      x                  0.0025 M              0.0010 M

Change                  0                       0                            0

Equilibrium             x                  0.0025 M              0.0010 M

K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}

Given that K_c = 1.4*10^2 ; Then:

1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}

1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}

3.5 =( \frac{(0.001)}{(x)})^2

\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}

1.87=\frac{(0.001)}{(x)}

(x)= \frac{(0.001)}{1.87 }

x = 5.35 *10^{-4}M

∴ The equilibrium concentration of CO = x = 5.35 *10^{-4}M

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