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Furkat [3]
2 years ago
13

A tiny sample of an aqueous solution of two substances R and P is sketched below as if it was under an imaginary microscope so p

owerful that molecules could be seen. (The water molecules are not shown.) R and P can interconvert. That is, R can turn into P, and P can turn back into R 2 The equilibrium constant K for this equilibrium is. Predict the number of R and P molecules in this sample when the interconversion reaches equilibrium. a number of R molecules: D number of P molecules:
Chemistry
1 answer:
PilotLPTM [1.2K]2 years ago
7 0

Answer:

R = 6, P = 6

Explanation:

Answer

Number of R molecules = 6

Number of P molecules = 6

Explanation

R(aq) <-------> P(aq)

K = [P]/[R]

at equillibrium

[ P ] = 10+x

[R] = 2-x

10 + x /(2-x) = 1

10 + x = 2- x

2x = -8

x = -4

Therefore,

[ R] = 2 - ( -4) = 6

[ P ] = 10 + ( -4) = 6

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Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:

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To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

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2 years ago
What is the millimolar solubility of oxygen gas, o2, in water at 16 ∘c, if the pressure of oxygen is 1.00 atm?
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Partial pressure is the amount of pressure or force that is exerted by the atoms into the outer environment. it is dependent on the temperature and pressure of the present surroundings. in this case, we are asked in this problem to determine the partial pressure of oxygen at 16oC and 1 atm. We have to look into a solubility data table commonly found in handbooks and determined via experiments and correlations. According to literature, the value of the partial pressure is equal to 0.617 mM.This is under the assumption that the salinity of the water in which oxygen is dissolved is equal to zero.
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2 years ago
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An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine
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Answer:

The molecular formula of the compound is C_{7}H_{6}O_{2}.

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

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m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)

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Hydrogen

m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)

m_{H} = 6.057\,g

Oxygen

m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)

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Now, the number of moles (n), measured in moles, of each element are calculated by the following expression:

n = \frac{m}{M}

Where:

m - Mass of the element, measured in grams.

M- Molar mass of the element, measured in grams per mol.

Carbon (m_{C} = 84.067\,g, M_{C} = 12.011\,\frac{g}{mol})

n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }

n = 7

Hydrogen (m_{H} = 6.057\,g, M_{H} = 1.008\,\frac{g}{mol})

n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }

n = 6

Oxygen (m_{O} = 31.995\,g, M_{O} = 15.999\,\frac{g}{mol})

n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }

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C_{7}H_{6}O_{2}

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