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Inessa [10]
3 years ago
14

A 1.5 M solution of NaOH was made in a laboratory. If the solution made had a volume of 4.5 L, how many grams of NaOH were added

?
Chemistry
1 answer:
Crank3 years ago
6 0

Answer:

270g

Explanation:

Given parameters:

Concentration of NaOH  = 1.5M

Volume  = 4.5L

Unknown

Mass of NaOH added  = ?

Solution:

To solve the problem, we need to find the number of moles of the NaOH first;

 Number of moles  = concentration x volume

 Number of moles  = 1.5 x 4.5  = 6.75mol  

Now;

  Mass  = Number of moles x molar mass

 Molar mass of NaOH  = 23 + 16 + 1  = 40g/mol

  Mass  = 6.75 x 40  = 270g

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A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati
Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

                                      =0.00465 \ Mol

\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

7 0
3 years ago
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