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uysha [10]
2 years ago
12

Match the solutions to the descriptions of the freezing points.a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O

b. One mole of the ionic compound CuSO4 dissolved in 1000 g H2O c. One mole of the nonelectrolyte C6H12O6 dissolved in 1000 g H2O1. Highest freezing point 2. Intermediate freezing point 3. Lowest freezing point
Chemistry
1 answer:
rosijanka [135]2 years ago
5 0

Explanation:

Depression in Freezing point

= Kf × i × m

where m is molality , i is Van't Hoff factor, m = molality

Since molality and Kf remain the same

depression in freezing point is proportional to i

i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2

i=1 for C2h6O

i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)

So the freezing point depression is highest for MgCl2 and lowest for C2H6O

so freezing point of the solution = freezing point of pure solvent- freezing point depression

since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point

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At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
How many moles are in 3.2x10^23 atoms of francium?
andrew11 [14]

Answer:

i do not know i think the answer is 23

Explanation:

6 0
3 years ago
The strength of an acid is affected by the polarity of the bond connected to the acidic hydrogen. The more highly polarized this
Shtirlitz [24]

Answer:

The answer is "\bold{HClO_3 > HClO_2 >HClO > HBrO}"

Explanation:

We arrange oxoacids to decrease the intensity of acids in this question. Or we may conclude all this from strongest to weakest acids they order oxoacids, that's why above given order is correct.

5 0
3 years ago
Give two possible sets of four quantum numbers for the electron in hydrogen when it is in its lowest energy state.
konstantin123 [22]

Answer:

Explanation:

The electron in the lowest energy state will be found in 1 s energy level.

set of 4 possible quantum numbers

Principal quantum no : n = 1 ,

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Magnetic quantum no m = 0

Spin quantum no  s = + 1/2

set of other quantum nos

Principal quantum no : n = 1 ,

Azimuthal quantum no l = 0

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4 0
3 years ago
Balance <br> Na2O+H2O → NaOH
AleksAgata [21]
Answer:
Na2O+H2O=2NaOH
Step by step exp.
Given:
Equation Na2O+H2O=NaOH
To find: Balance the equation
Solution:
Taking LHS of the equation
LHS=Na2O+H2O
There is 2 sodium, 2 oxygen,& 2 hydrogen
To balance the equation we have equal number of atom so we multply 2 to the RHS=2NaOH
There fore the equation form is
Na2O+H2O=2NaOH
8 0
3 years ago
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