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Damm [24]
3 years ago
14

How many moles of nacl are required to make 250 ml of a 3.00 m solution?

Chemistry
1 answer:
choli [55]3 years ago
4 0
The molarity is calculated using the following rule:
molarity = number of moles of solvent / volume of solution (in liters)

We have the volume of solution = 250 ml = 0.25 liters and the molarity = 3 m

Substituting in the equation, we get:
3 = number of moles / 0.25
number of moles = 3 x 0.25 = 0.75 moles
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Chemists can use moles to calculate: A. How much of the products are needed and how much reactant will be made. B. How much prod
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Answer:

c.- How much of the reactants are needed and how much product will made.

Explanation:

The moles is the matter unit used in chemistry to simplify  some calculations, instead of using grams. Also the moles are very useful because the chemical reaction can be balanced.

When a Chemical reaction is balanced, then it can be easily to calculate how many moles are necessary to add in a process to obtain a quantity of grams of a product.

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The aldol reaction is catalyzed by acid as well as base. What is the reactive nucleophile in the acid-catalyzed aldol reaction
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Answer:

The reactive nucleophile is Ketone.

Explanation:

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A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th
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The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.  

The mass percent of lithium hydroxide can be calculated with the following equation:  

\% = \frac{m_{LiOH}}{m_{t}} \times 100    (1)

Where:

m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g   (2)  

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.    

\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}

0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}

\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    (3)                                        

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:  

\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    

\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol              

Solving for m_{LiOH}, we have:

m_{LiOH} = 0.082 g

Hence, the percent lithium hydroxide is (eq 1):

\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%  

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

  • brainly.com/question/6992535?referrer=searchResults
  • brainly.com/question/5840377?referrer=searchResults

I hope it helps you!                        

5 0
2 years ago
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