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Fudgin [204]
3 years ago
6

Choose the aqueous solution below with the highest boiling point. These are all solutions of nonvolatile solutes and you should

assume ideal van't Hoff factors where applicable.
Chemistry
1 answer:
o-na [289]3 years ago
8 0

The question is incomplete, the complete question is;

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.

Answer:

AlCl3 0.100 m

Explanation:

Let us remember that the boiling point elevation is given by;

ΔTb = Kb m i

Where;

ΔTb = boiling point elevation

Kb = boiling point constant

m = molality of the solution

i = Van't Hoff factor

We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.

AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.

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1. 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? (Specific heat capacity of Cu is 0.092 cal
iren2701 [21]

 The amount  of energy  that  was  used to heat Cu  is 27.6 cal

 

<u><em>calculation</em></u>

Heat (Q) = M ( mass)  x c(specific heat capacity)  x ΔT( change in temperature)

where;

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Q is therefore = 5.0 g  x 0.092 cal / g°c x 60°c =27.6 cal

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4 years ago
The energy of motion is called
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3 years ago
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How many molecules does 88 grams CO2 contain? A) 1.2 X 1024 B) 1.8 X 1024 C) 3.0 X 1023 D) 6.0 X 1023
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Hey there!:

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7 0
3 years ago
A sample of an unknown gas takes 222 s to diffuse through a porous plug at a given temperature. At the same temperature, N2(g) t
gulaghasi [49]

Answer:

\large \boxed{\text{45.1 g/mol}}

Explanation:

Graham’s Law applies to the diffusion of gases:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases, the ratio of their rates of diffusion is

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\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}

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t₂ = 222 s

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Calculation :

\begin{array}{rcl}\dfrac{222}{175} & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.269 & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.609 & = & \dfrac{M_{2}}{28.01}\\\\M_{2} & = & 1.609 \times 28.01\\ & = & \textbf{45.1 g/mol}\\\end{array}\\\text{The molar mass of the unknown gas is $\large \boxed{\textbf{45.1 g/mol}}$}

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