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PolarNik [594]
3 years ago
8

g A and B are both non-volatile molecular compounds. Substance A has a greater molar mass than substance B. If 50 g of substance

A are dissolved in 250 g of water in one beaker, and 50 g of substance B are dissolved in 250 g of water in another beaker, then which ONE statement below will be true? A) the vapor pressure of solution A will be lower than the vapor pressure of solution B. B) the solution of A will freeze at a lower temperature than the solution of B. C) the two solutions will have the same vapor pressure. D) the boiling point of solution A will be lower than the boiling point of solution B. E) the solution of A will have a higher osmotic pressure than the solution of B.
Chemistry
1 answer:
Doss [256]3 years ago
8 0

Answer:

The correct answer is - option D.  the boiling point of solution A will be lower than the boiling point of solution B

Explanation:

Colligative properties such as a decrease in the freezing point of the solution, increase in the boiling point of substance, decrease in Lowering of vapor pressure, and other properties depend upon the number of molecules only.

In the given solution the equal amount of two solutions are mixed that is 50 grams however due to the difference in the molecular mass so the atoms present in both solution A and B will be different. It is known that the number of atoms of a substance is inversely proportional to the molecular mass of the particular substance.

As it is given that Solution B has a low molecular mass which means it has a high number of atoms that means its boiling point will be higher than solution A.

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Consider the following reaction where Kc = 1.80×10-2 at 698 K:
Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

Qc=\frac{[C]^{c}*[D]^{d}  } {[A]^{a}*[B]^{b}}

In this case:

Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

  • [H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}=2.09*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}=4.14*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{0.280 moles}{1 Liter}= 0.280 \frac{moles}{liter}

So,

Qc=\frac{2.09*10^{-2} *4.14*10^{-2}  } {0.280^{2} }

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0
3 years ago
Solve for Va<br><br>MaVa=MbVb​
PolarNik [594]

Answer:

Va = (MbVb)/Ma

Explanation:

Divide both sides by Ma and voila!

8 0
3 years ago
Mass, size, and color are examples of what?
Igoryamba
Physical properties of matter
3 0
3 years ago
What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?
Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

6 0
3 years ago
Ions are atoms that have gained or lost which subatomic particles?
Bad White [126]
Protons cannot be lost without the atoms becoming an entirely different element. Elements can infact have different numbers of neutrons within the same element, but neutrons are not related to the electrical charge. The answer is c. electrons.
7 0
3 years ago
Read 2 more answers
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