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3 years ago

How many grams are in 2.06x10-4 moles of calcium phosphate

Chemistry

1 answer:

Orlov [11]3 years ago

3 0

Answer:

63.86 mg

Explanation:

Molar mass is the mass of 1 mol of compound / atom.

molar mass of the compound is - 310 g/mol

mass of 1 mol is - 310 g

therefore mass of 2.06 x 10⁻⁴ mol is - 310 g/mol x 2.06 x 10⁻⁴ mol = 0.06386 g

mass of calcium phosphate is - 63.86 mg

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What is the formula ofchromium(III) hydrogensulfate?

FromTheMoon [43]

Answer:

Cr (HSO4)3

Explanation:

its molecular weight is 343.20 g/mol

its molecular formula can also be written as CrH3O12S3

molar mass of Cr (HSO4)3 can be calculated by following method;

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3 0

3 years ago

Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

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3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 60.5 kJ/mol at 600. K by the introduction of a catalyst. If t

Ahat [919]

Answer:

The catalyzed reaction will take 1,41 s

Explanation:

The rate constant for a reaction is:

$k = A e^{-\frac{Ea}{RT}}$

Assuming frequency factor is the same for both reactions (with and without catalyst) it is possible to obtain:

${\frac{k1}{k2}} = e^{-\frac{Ea_{2}-Ea_{1}}{RT}}$

Replacing:

${\frac{k1}{k2}} = e^{-\frac{60,5kJ/mol-184kJ/mol}{8,314472x10^{-3}kJ/molK*600k}}$

${\frac{k1}{k2}} = 5,64x10^{10}$

That means the reaction occurs 5,64x10¹⁰ faster than the uncatalyzed reaction, that is 2537 years / 5,64x10¹⁰ = 4,50x10⁻⁸ years. In seconds:

4,50x10⁻⁸ years× $\frac{365days}{1year}$ × $\frac{24hours}{1day}$ × $\frac{3600s}{1hour}$ =<em> 1,41 s</em>

I hope it helps!

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3 years ago