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fenix001 [56]
3 years ago
14

How many moles are in 50 grams of cobalt? help asapp

Chemistry
1 answer:
sammy [17]3 years ago
3 0

hey there!

atomic mass cobalt = 58.9332 amu

therefore:

1 mol Co ------------ 58.9332 g

moles ------------ 50 g

moles = 50 x 1 / 58.9332

moles  = 50 / 58.9332

 = 0.8484 moles of Co

Hope this helps!

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1) Describe the characteristics of molecular compounds that affect their particular physical
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The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.
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Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a
horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound            Amount

     Ni                        12.7 g

   CO                        1.98 g

Ni(CO)_4                  0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of Ni(CO)_4 = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M

\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M

\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M

For the given chemical reaction:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

The expression of equilibrium constant for the reaction:

K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0
4 years ago
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charle [14.2K]

Answer:

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Explanation:

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i guess its e) Mn (VII)

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6 0
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