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vovangra [49]
2 years ago
15

A beaker holds 962 g of a brine solution that is 6.20 percent salt. If 123g of water are evaporated from the beaker, how much sa

lt must be added to have an 8.60 percent brine solution? How many grams of the 8.6% brine solution will be produced?
Chemistry
1 answer:
denpristay [2]2 years ago
8 0

Answer:

13.687 grams of salt should be added

The total grams of 8.6% of brine solution produced is 852.687g

Explanation:

Solution mass= 962g

Salt= 6.2%

Water = 93.8%

962 gram of water is made up of:

902.356g ( due to vaporization which reduces mass)

= 902.356 - 123

= 779. 356g of water

59.644g of salt.

If we add x gram of salt for making brine solution up to 8.6%

=(59. 644g + x)g.of salt

% salt = Mass of Salt / Total mass of solution

= 0.086= 59.644 + x / 779.356 + 59.644 + x

= 59.644 + x / 839 + x

x= 13.687 g of salt

Grams of 8.6% brine solution will be:

Gram of water + total gram of salt added to form 8.6% brine solution.

= 779.356g +59.644g + 13.687g

= 852.687g

The total grams of 8.6% of brine solution produced is 852.687g

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The correct answer is option d, that is, atoms of the element.  

As the atoms are neither destroyed nor created in a chemical reaction, the sum of the mass of the products in a reaction must be equivalent to the sum of the mass of the reactants.  

The chemical reactions must be balanced, they must exhibit a similar number of atoms of each element on both the sides of the equation. As a consequence, the mass of the reactants must be equivalent to the mass of the products of the reaction.  


7 0
2 years ago
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mixture or pure substance: 1.blood 2.dyes 3.self-raising flour 4.muesli 5.copper wire 6.distilled water 7.table salt 8.milk 9.br
jenyasd209 [6]

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Pure substance: Copper wire, distilled water, table salt, oxygen.

Explanation:

Mixture is a substance which is made up two or more number of compounds which chemically inactive and retain their distinct chemical properties.

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Pure substance is defined as anything with uniform and unchanging composition is known s pure substance.

Copper wire, distilled water, table salt, oxygen.

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A pound is approximately 0.45 kilogram. A person weighs 87 kilograms. What is the person’s weight, in pounds, when expressed to
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A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t
TEA [102]

Explanation:

The reaction equation will be as follows.

     Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)

So, according to this equation, 1 mole Ca(OH)_{2} = 2 mol HBr = 1 mol CaBr_{2}

Therefore, calculate the number of moles of calcium hydroxide as follows.

     No. of moles of Ca(OH)_{2} = V \times Molarity

                                    = 50 \times 0.6

                                    = 30 mmol

Similarly, calculate the number of moles of HBr as follows.

        No. of moles of HBr = M \times V

                                          = 50 \times 0.6

                                          = 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of CaBr_{2} = 30 \times \frac{1}{2}

                                                     = 15 mmol

Hence, calculate the amount of heat released as follows.

                Heat released in the reaction(q) = m \times s \times \Delta T

as,    m = mass of solution

and,             Density = \frac{mass}{volume}

or,                  mass = Density × Volume

                               = 1.08 g/ml \times (50 + 50) ml

                               = 108 g

where,    s = specific heat of solution = 4.18 j/g.k

and,        change in temperature \Delta T = (26 - 23)^{o}C

                                                                 = 3
^{o}C

Hence, the heat released will be as follows.

                   q = m \times s \times \Delta T

                        q = 108 \times 4.18 \times 3^{o}C

                           = 1354.32 joule

or,                        = 1.354 kJ       (as 1 kJ = 1000 J)    

Also,          \Delta H_{rxn} = \frac{-q}{n}

                              = \frac{-1.354}{15 \times 10^{-3}}

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Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

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2 years ago
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