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katen-ka-za [31]
3 years ago
15

A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found

to have a melting point of instead. Calculate the molal freezing point depression constant of . Round your answer to significant digits.
Chemistry
1 answer:
pshichka [43]3 years ago
5 0

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for <em>freezing point depression</em>:

  • ΔTf = Kf * m

In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

  • 350 g X ⇒ 350/1000 = 0.35 kg X
  • 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea

Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to <u>solve for Kf</u>:

  • ΔTf = Kf * m
  • 3.3 °C = Kf * 1.51 m
  • Kf = 2.2 °C/m
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3 years ago
While mercury is very useful in barometers, mercury vapor is toxic. Given that mercury has a ΔHvap of 59.11 kJ/mol and its norma
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2.38×10^-3

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8 0
3 years ago
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =
Lady bird [3.3K]

Answer:

4.86\times10^{-7}\ \text{m}

Explanation:

R = Rydberg constant = 1.09677583\times 10^7\ \text{m}^{-1}

n_1 = Principal quantum number of an energy level = 2

n_2 = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}

The wavelength of the light emitted is 4.86\times10^{-7}\ \text{m}.

6 0
3 years ago
For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.
Ira Lisetskai [31]

Explanation:

The reaction is given as;

Br2(g) + Cl2(g) ----> 2BrCl(g)

From the equation;

1 mol of Br2 reacts with 1 mol of Cl2

Converting the masses given to moles, using the formular;

Number of moles = Mass / Molar mass

Br2;

Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol

Cl2;

Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol

From the values;

0.18334 mol of Cl2 would react with 0.18334 mol of Br2 with an excess of 0.02754 mol of Br2

<em>What is the maximum amount of bromine monochloride that can be formed? __________grams</em>

<em />

1 mol of Cl2 produces 2 mol of Bromine Monochloride

0.18334 mol of Cl2 would produce x

Solving for x;

x = 0.18334 * 2 = 0.36668 mol

Converting to mass;

Mass = Number of moles * Molar mass = 0.36668 mol * 115.357 g/mol

Mass = 42.299 g

<em />

<em>What is the FORMULA for the limiting reagent?</em>

<em />

The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.

<em>What amount of the excess reagent remains after the reaction is complete? __________grams</em>

The excess reagent is Br2

The number of moles left is;

0.02754 mol of Br2

Converting to mass;

Mass = Number of moles * Molar mass = 0.02754 mol  * 159.808 g/mol

Mass = 4.401 g

7 0
2 years ago
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