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3 years ago

[AZ][B]^4=K

Chemistry

1 answer:

elena-s [515]3 years ago

4 0

Answer:

I think D. shift the equilibrium reaction to favor the endothermic process

Explanation:

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4 years ago

The fossils in line 4 __ are the ones in line 1

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In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the

N76 [4]

The question is incomplete, here is the complete question:

Consider the following reaction: $2N_2O(g)\rightarrow 2N_2(g)+O_2(g)$

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

<u>Answer:</u> The average rate of appearance of oxygen gas is $2.64\times 10^{-3}M/s$

<u>Explanation:</u>

We are given:

Moles of oxygen gas = $1.9\times 10^{-2}moles$

Volume of solution = 0.480 L

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

So, $\text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M$

The given chemical reaction follows:

$2N_2O(g)\rightarrow 2N_2(g)+O_2(g)$

The average rate of the reaction for appearance of $O_2$ is given as:

$\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}$

Or,

$\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}$

where,

$C_2$ = final concentration of oxygen gas = 0.0396 M

$C_1$ = initial concentration of oxygen gas = 0 M

$t_2$ = final time = 15.0 s

$t_1$ = initial time = 0 s

Putting values in above equation, we get:

$\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s$

Hence, the average rate of appearance of oxygen gas is $2.64\times 10^{-3}M/s$

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4 years ago

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3 years ago

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How many atoms of Lead are in a 62.7 g sample of Lead(II) permanganate?

Alexandra [31]

Answer:

$\large \boxed {8.48 \times 10^{22}\text{ atoms}}$

Explanation:

1 mol of Pb(MnO₄)₂contains 1 mol of Pb atoms.

1. Moles of Pb

$\text{Moles of Pb} = \text{62.7 g Pb(MnO$_{4}$)}_{2} \times \dfrac{\text{1 mol Pb(MnO$_{4}$)}_{2}}{\text{445.07 g Pb(MnO$_{4}$)}_{2}}\\\\\times \dfrac{\text{1 mol Pb}}{\text{1 mol Pb(MnO$_{4}$)}_{2}} = \text{0.1409 mol Pb}$

2. Atoms of Pb

$\text{Atoms of Pb} = \text{0.1409 mol Pb } \times \dfrac{ 6.022 \times 10^{23}\text{ atoms Pb }}{\text{1 mol Pb }}\\\\= \large \boxed {\mathbf{8.48 \times 10^{22}}\textbf{ atoms Pb }}$

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3 years ago