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Snezhnost [94]
3 years ago
13

If the reaction is at dynamic equilibrium at 500 K, which statement applies to the given chemical system?

Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0

The correct option is this: THE CONCENTRATION OF THE PRODUCTS AND THE REACTANTS DO NOT CHANGE.

A reversible chemical reaction is said to be in equilibrium if the rate of forward reaction is equal to the rate of backward reaction. At this stage, the concentrations of the products and the reactants remain constant, that is, there is no net change in the concentration even though the reacting species are moving between the forward and the backward reaction.

Arte-miy333 [17]3 years ago
7 0
If the reaction is at dynamic equilibrium at 500 K, the statement that applies to the given chemical system is <span>The concentrations of the products and reactants do not change.</span>
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4 0
3 years ago
A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o
Andre45 [30]

Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

Regards.

7 0
3 years ago
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