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Shalnov [3]
3 years ago
9

A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/ml. what mass of the salt in grams is in 5.00l of

this solution?
Chemistry
1 answer:
meriva3 years ago
6 0
To answer this item, we solve first for the mass of the solution by multiplying the density by the volume. That is,

      m = (density)(volume)

Substituting the known values,
    m = (1.50 g/mL)(5L)(1000 mL/1L)
      m = 7500 grams

To determine the mass of the salt in the solution, multiply the calculated mass of the solution by the decimal equivalent of the percent salt in the solution.

      m of salt = (7500 g)(0.33)
      m of salt = 2475 grams

<em>Answer: 2475 grams</em>
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Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p
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Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

\text{Rate of formation of }H=+\frac{d[H]}{dt}

\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

Given:

-\frac{d[D]}{dt}=0.18mol/L.s

As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

+\frac{d[H]}{dt}=2\times 0.18mol/L.s

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We know that molarity is moles of solute per liter of solution.

If molarity and volume is given then, moles of solute is molarity times volume in liters.

moles of solute = molarity* liters of solution

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