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Andreas93 [3]
3 years ago
14

What volume (in liters) of a solution contains 0.14 mol of KCl?

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
5 0

Answer:

\boxed {\boxed {\sf 0.078 \ L }}

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

  • moles of solute = 0.14 mol KCl
  • molarity= 1.8 mol KCl/ L
  • liters of solution=x

Substitute these values/variables into the formula.

1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}

1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl

1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

The units of moles of potassium chloride cancel.

x= \frac{0.14 }{1.8 L}

x=0.07777777778 \ L

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

x \approx 0.078 \ L

There are approximately <u>0.078 liters of solution.</u>

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Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

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Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

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3 0
3 years ago
Calculate the enthalpy of the reaction
harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)     \Delta H_A=+2035kJ

(2) 2B(s)+3H_2(g)\rightarrow B_2H_6(g)    \Delta H_B=+36kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_C=-285kJ

(4) H_2O(l)\rightarrow H_2O(g)    \Delta H_D=+44kJ

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D

\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)

\Delta H=-2552kJ

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0
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