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3 years ago

2. Calculate the molarity of a solution in which 0.50 moles of MgCl2 are dissolved to produce 1.5 liters

Chemistry

1 answer:

vodka [1.7K]3 years ago

6 0

Answer:

Molarity= 0.33M

Explanation:

From n= CV

n= 0.5mol, V= 1.5 ,C = ?

Substitute into above formula

0.5= C×1.5

C= 0.33M

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What is the formula for manganese (ii) fluoride decahydrate? (you may use a * to represent the dot in the formula.)?

Rasek [7]

These types of molecules are called hydrates. They have a certain number of moles attached to the salt. Their characteristic is being hygroscopic. That means that when they are exposed to air, they readily solvate.

The formula for Manganese Fluoride Decahydrate will involve the formula Mn, F and H₂O. In ionic form, Manganese is Mn⁺² while fluoride is in F⁻. When they are brought together, their superscripts are 'cross-multiplied' and becomes their respective subscripts. The compound becomes MnF₂. Then, we add the decahydrate which means 10 moles of H₂O. Hence, the formula for Manganese Fluoride Decahydrate is MnF₂*10H₂O.

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Hence option B is correct.

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3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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tino4ka555 [31]

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