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yKpoI14uk [10]
4 years ago
8

What molality of a nonvolatile, nonelectrolyte solute is needed to raise the boiling point of water by 7.10°c (kb = 0.520°c/m)?

Chemistry
1 answer:
faltersainse [42]4 years ago
7 0
According to boiling point elevation equation:
Δ T = i Kb m
when ΔT (change in boiling point) = 7.10 C° 
and i (van't Hoff factor)= 1
and Kb = 0.520
so, by substitution:
7.10 = 1*0.520 *m
m = 7.1 / 0.52 = 13.65 m
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NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3?
ad-work [718]
<h2>NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3? </h2><h2> </h2><h2>A. It has a very low pH. </h2><h2>B. It's dissociation is a reversible reaction. </h2><h2>C. It has a high H+ concentration. </h2><h2>D. It will release all of its OH- ions.</h2>

Explanation:

<h3>NH3 is a weak alkali that does not dissociate fully into its solution: It's dissociation is a reversible reaction. </h3><h3></h3>

Reactions are also :

  • Reversible
  • Irreversible

Reversible reaction

A reaction in which products can combine back to give reactants under same given condition .

Example : N₂+H₂-------NH₃

Irreversible reaction

A reaction in which the products cant combine back to give reactants under same set of conditions .

Example : Burning of paper

3 0
3 years ago
Read 2 more answers
Glacier National Park in Montana is approximately 4,100 ft above sea level with an atmospheric pressure of 681 torr. At what tem
lapo4ka [179]

Answer : The temperature of liquid is, 369.9 K

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of liquid at 373 K = 681 torr

P_2 = vapor pressure of liquid at normal boiling point = 760 torr

T_1 = temperature of liquid = ?

T_2 = normal boiling point of liquid = 373 K

\Delta H_{vap} = heat of vaporization = 40.7 kJ/mole = 40700 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760torr}{681torr})=\frac{40700J/mole}{8.314J/K.mole}\times (\frac{1}{T_1}-\frac{1}{373K})

T_1=369.907K\approx 369.9K

Hence, the temperature of liquid is, 369.9 K

6 0
3 years ago
Read 2 more answers
SOMEONE PLEASE HELP! WILL GIVE BRAINLIEST!
Elis [28]

Answer:

1 At 0C° KNO3 is least soluble

2 Approximately 65 grams

3 About 30 grams

4 yes it increases at the same rate can be explained by straight line graph

Explanation:

3 0
2 years ago
What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

4 0
3 years ago
Need help asap<br> will give brainly
uranmaximum [27]

Answer:

8. B

9. D

10. A

Explanation:

give brainliest pls ^-^

4 0
2 years ago
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