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dmitriy555 [2]
2 years ago
11

The mass percent of solute in a solution containing 3.73 g KBr dissolved in 131 g of H2O is:

Chemistry
1 answer:
Elina [12.6K]2 years ago
7 0

\huge \underbrace \mathfrak \red{Answer}

28%

Explanation:

mass of solute(KBr) = 3.73g

mass of solvent(H2O) = 131g

mass of solution = mass of solute + mass of solvent

= 3.73 + 131

= 134.73g

\sf \large {mass \: percentage =  \frac{mass \: of \: solute}{mass \: of \: solvent}  \times 100} \\  \\  \sf  mass \: percentage =  \frac{3.73}{134.73}  \times 100 \\  \\  \sf mass \: percentage =  0.028 \times 100 \\  \\  \sf mass \: percentage = 28\%

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oksano4ka [1.4K]

Answer:

hope this helps :)

Explanation:

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5 0
2 years ago
Read 2 more answers
A clear liquid in an open container is allowed to evaporate. After three days, a solid is left in the container. Was the clear
Mashutka [201]

The liquid did not chemically bond after 3 days, therefor it is a mixture.

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7 0
3 years ago
A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

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b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
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sdas [7]

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3 years ago
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Answer:

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