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3 years ago

5

A compound known to contain only molybdenum (Mo) and oxygen (O) is found to be 67% molybdenum by mass What is the empirical form

ula of the compound?

1 answer:

Eddi Din [679]3 years ago

4 0

Empirical formula: MoO3

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Which activity might help to increase the validity of this experiment?

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Given that a theoretical yield for isolating Calcium Carbonate in this experiment would be 100%. From that information and based

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It's well Explained below.

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3 years ago

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Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

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Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

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$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

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

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

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

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

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Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

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