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liubo4ka [24]
3 years ago
10

1 Point

Chemistry
1 answer:
KatRina [158]3 years ago
6 0
D removing thermal energy from the water
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25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

5 0
3 years ago
What’s molar mass? For chemistry
Alik [6]
In chemistry, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles.
3 0
3 years ago
Arrange the following compounds from left to right in order of increasing percentage by mass of hydrogen: a) H2O, b) C12H26, c)
Nuetrik [128]

Answer:

(A) N4H6  (B) H2O  (C) LiH  (D) C12H26

Explanation:

The given compounds have been arranged  from left to right in order of increasing percentage by mass of hydrogen.

The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.

4 0
3 years ago
If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees
telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁  +  q₂  =  0

n  =  moles of NH₄NO₃  =  8.00 g NH₄NO₃  ×  1 mol NH₄NO₃/80.0 g NH₄NO₃          

∴ n =   0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m  =  mass of solution  =  1000.0 g + 8.00 g  =  1008.0 g

q₂  =  mcΔT  = 58.0 g  ×  4.184 J°C⁻¹  g⁻¹  × ((20.39-21)°C) = -2570.19 J

q₁  +  q₂  =  0.100 mol  ×ΔHsoln  – 2570.19 J  =  0

ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

7 0
3 years ago
What are the coefficients that would balance this equation _____NaOH+____H2CO3-Na2CO3+_____H2O
FromTheMoon [43]
I don’t see any equal signs to make it an equation. Am I missing something?
4 0
2 years ago
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