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slega [8]
3 years ago
7

You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.

Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Answer:

Solution is 4.67% by mass of salt

Explanation:

% by mass is the concentration that defines the mass of solute in 100g of solution.

In this case we have to find out the mass of solution with the data given:

Mass of solution = Mass of solute + Mass of solvent

Solute:  Salt → 14.2 g

Solvent: Water → 290 g

Solution's mass = 14.2 g + 290g = 304.2 g

% by mass = (mass of solute / mass of solution) . 100

(14.2 g / 304.2g) . 100 = 4.67 %

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If you have 1.29 moles of NaCl in 4 liters of solution, what is the molarity of the solution?
Illusion [34]

Answer:

0.32M

Explanation:

Given parameters:

Number of moles of NaCl  = 1.29moles

Volume of solution = 4liters  = 4dm³

Unknown:

Molarity of solution = ?

Solution:

Molarity is one the ways of expressing the concentration of a solute in a solution. It is given as;

         Molarity  = \frac{number of moles }{volume}

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           Molarity  = \frac{1.29}{4}    = 0.32mol/dm³ or 0.32M

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2 years ago
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astra-53 [7]

Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

2SO3 —> O2 + 2SO2 (196 kJ)

Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

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Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

S + O2 —> SO2

Now, we must add the enthalpies together to get our final answer.

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Hope this helps!

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