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Elenna [48]
1 year ago
6

How many grams of mgcl2 is produced if 4. 67 moles of hcl react

Chemistry
1 answer:
frosja888 [35]1 year ago
7 0

The mass of MgCl₂ produced is 5.685 grams.

<h3>What is mass?</h3>

Mass is the quantity of matter of a physical body.

The reaction is

Mg   +   2HCl   →    MgCl₂ + H₂

2 mole of HCl

(2 × 36.458) = 72.8 g        

                   

1 moles of MgCl₂

(1 × 95.21 g) = 95.21 g

1 g  Mg will produce = \dfrac{95. 21}{78.2}

4.67 mol Mg will produce

\dfrac{95. 21}{78.2} \times 4.67 = 5.685

Thus, the mass of MgCl₂ produced is 5.685 grams.

Learn more about mass

brainly.com/question/19694949

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Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
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Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

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49.9 ml of a 0.00292 m stock solution of a certain dye is ddiluted to 1.00 L. the diluted solution has an absorbance of 0.600. w
inna [77]

Complete Question

49.9 ml of a 0.00292 m stock solution of a certain dye is diluted to 1.00 L. the diluted solution has an absorbance of 0.600. what is the molar absorptivity coefficient of the dye

Answer:

The  value is  \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

Explanation:

From the question we are told that

   The volume of the stock solution is  V_1   =  49.9 mL  =  0.0499 \  L  

   The concentration of the stock solution is  C_1  =  0.00292 \  M

   The volume of the diluted solution is  V_2 =  1.00 \  L

   The absorbance is  A =  0.600

Generally the from the titration equation we have that

         C_1 * V_1 =  C_2 * V_2

=>      0.00292  * 0.0499 =  C_2 * 1

=>     C_2 = 0.0001457 \  M

Generally from  Beer's law we have that

      A  =  \epsilon  * l  *  C_2

=>   \epsilon  =  \frac{A}{ l  *  C_2 }

Here  l is the length who value is  1 cm because the unit of  molar  absorptivity coefficient of the dye is M^{-1} *  cm^{-1}

So

            \epsilon  =  \frac{0.600}{ 1   * 0.0001457   }  

=>       \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

4 0
2 years ago
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