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Nadya [2.5K]
3 years ago
8

A beaker contains 20.0 ml of a 50.0 g/L solution. What will the new concentration be if you add 40.0 mL of water?

Chemistry
1 answer:
Ronch [10]3 years ago
8 0

Answer:

moles SO42- = 0.0500 L x 0.20 M=0.010

moles Ba2+ = 0.0500 L x 0.10 M = 0.0050

Ba2+ + SO42- = BaSO4 (s)

moles SO42- in excess = 0.010 - 0.0050=0.0050

total volume = 100 mL = 0.100 L

[SO42-]= 0.0050/0.100= 0.050 M

[Na2SO4] = 2 /2 = 1 M

moles Na2SO4 = 2 M x 0.500 L = 0.500

mass Na2SO4 = 0.500 x 142 g/mol=71.0 g

moles MgBr2 = 46 /184 =0.25

moles Br- = 0.25 x 2 = 0.50

[Br-]= 0.50 / 0.50 L = 1 M

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Answer:

T_f = 25.05°C

Explanation:

Given:

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since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

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number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

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= 21.84 kJ

∴ Heat absorbed by the calorimeter =  C_{calorimeter} (T_f-T_i} )

21.84 kJ =6.15 kJ/°C * (T_f-21,5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - 132.225 kJ

21.84 KJ + 132.225 kJ = (6.15 kJ/^0C * T_f)

154.065 kJ = (6.15 kJ/^0C * T_f)

T_f = \frac{154.065kJ}{6.15kJ/^0C}

T_f =25.05°C

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