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Paul [167]
3 years ago
14

Gas y effuses half as fast as o2. what is the molar mass of gas y?

Chemistry
1 answer:
devlian [24]3 years ago
4 0
To solve this question, we will use Graham's law which states that:
(R1 / R2) ^ 2 = M2 / M1 where
R1 and R2 are the rates of effusion and M1 and M2 are the molar masses of the two gases.
From the periodic table, we can calculate the molar mass of O2 as follows:
molar mass of O2 = 2*16 = 32 grams

Therefore we have:
R1 / R2 = Ry / RO2 = 1/2
M1 is My we want to get
M2 is molar mass of O2 = 32 grams
Substitute in the above equation to get the molar mass of y as follows:
(1/2) ^2 = (32/My)
1/4 = 32/My
My = 32*4 = 128

Therefore, molar mass of gas y = 128 grams

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1. If a solution of sodium chloride has 22.3 g of
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<u>We are given:</u>

Mass of NaCl in the given solution = 22.3 grams

Volume of the given solution = 2 L

<u></u>

<u>Number of Moles of NaCl:</u>

We know that the number of moles = Given mass / Molar mass

Number of moles = 22.3 / 58.44 = 0.382 moles

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<u>Molarity of NaCl in the Given solution:</u>

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<em />

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A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis o
MAVERICK [17]

Answer : The value of \Delta G_{rxn} is -49.6 kJ/mol

Explanation :

First we have to calculate the reaction quotient.

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

ATP(aq)+H_2O(l)\rightarrow ADP(aq)+HPO_4^{2-}(aq)

The expression for reaction quotient will be :

Q=\frac{[ADP][HPO_4^{2-}]}{[ATP]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Given:

[ATP] = 5.0 mM

[ADP] = 0.60 mM

[HPO_4^{2-}] = 5.0 mM

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

Q=\frac{(0.60)\times (5.0)}{(5.0)}=0.60mM=0.60\times 10^{-3}M

Now we have to calculate the value of \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient = 0.60\times 10^{-3}

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-30.5kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (310K)\times \ln (0.60\times 10^{-3})

\Delta G_{rxn}=-49.6kJ/mol

Therefore, the value of \Delta G_{rxn} is -49.6 kJ/mol

6 0
3 years ago
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