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3 years ago

The normal freezing point of a certain liquid

Chemistry

1 answer:

slavikrds [6]3 years ago

3 0

Answer : The molal freezing point depression constant of liquid X is, $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X = $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

$0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}$

$K_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of liquid X is, $4.12^oC/m$

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A 0.327-g sample of azulene (C10H8) is burned in a bomb calorimeter and the temperature increases from 25.20 °C to 27.60 °C. The

liq [111]

Explanation:

The given data is as follows.

Molecular weight of azulene = 128 g/mol

Hence, calculate the number of moles as follows.

No. of moles = $\frac{mass}{\text{molecular weight}}$

= $\frac{0.392 g}{128 g/mol}$

= 0.0030625 mol of azulene

Also, $-Q_{rxn} = Q_{solution} + Q_{cal}$

$Q_{rxn} = n \times dE$

$Q_{solution} = m \times C \times (T_{f} - T_{i})$

$Q_{cal} = C_{cal} \times (T_{f} - T_{i})$

Now, putting the given values as follows.

$Q_{solution} = 1.17 \times 10^{3} g \times 10^{3} \times 4.184 J/g^{o}C \times (27.60 - 25.20)^{o}C$

= 11748.67 J

So, $Q_{cal} = 786 J/^{o}C \times (27.60 - 25.20)^{o}C$

= 1886.4 J

Therefore, heat of reaction will be calculated as follows.

$-Q_{rxn}$ = (11748.67 + 1886.4) J

= 13635.07 J

As, $Q_{rxn} = n \times dE$

13635.07 J = $-n \times dE$

dE = $\frac{13635.07 J}{0.0030625 mol}$

= 4452267.75 J/mol

or, = 4452.26 kJ/mol (as 1 kJ = 1000 J)

Thus, we can conclude that $\Delta E$ for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.

7 0

3 years ago

Which option identifies all the bonds broken during this chemical reaction?

Tresset [83]

Answer:

the bond b/w oxygen and hydrogen atom

7 0

2 years ago

Read 2 more answers

i don't need these solved I just need the steps to solve it and. what these problems in chemistry would be called so I can look

wolverine [178]

The answers are as follows,

1. 4.71 L,

2. 3.29 L,

3. 1634.6 torr

<u>Explanation:</u>

All the above problems can be sorted out using the Boyle's law represents the relationship between the volume and pressure.

It can be expressed as follows,

P1V1 = P2V2

Rearranging the above expression, we get,

$$ V2 = \frac{P1V1}{V2}$

1) The first case can be solved as,

$$V 2=\frac{921 \mathrm{mm} \mathrm{Hg} \times 3.89 \mathrm{L}}{760 \mathrm{mm} \mathrm{Hg}}=4.71 \mathrm{L}$

2) The second case can be solved as,

$$V 2=\frac{25 L \times 600 \text { torr}}{4560 \text { torr}}=3.29 L$

3) The third case can be solved as,

$$P 2=\frac{7.43 L \times 1100 \text { torr}}{5 L}=1634.6 \text { torr}$

Thus the answers for all the three cases are found as,

1. 4.71 L,

2. 3.29 L,

3. 1634.6 torr respectively

6 0

2 years ago

2. What is the molarity of a HCl solution containing 1.2 grams of hydrochloric acid in 0.025 L of solution?

alisha [4.7K]

Answer:

1.3M

Explanation:

Convert from grams to moles:

molar mass of HCl = 1.01g(molar mass of H) + 35.45g(molar mass of Cl) = 36.46g HCl

1.2g HCl (1mol HCl/36.46g HCl)

= .0329 mol HCl

Molarity = mol/L (important formula for concentration)

Plug your values in:

Molarity = .0329mol/.025L

1.317M - but you used two significant figures in the question, so:

1.3M

3 0

3 years ago

Enter the appropriate symbol for an isotope of phosphorus-32 corresponding to the isotope notation AZX. Express your answer as a

Lelechka [254]

Answer:

$^{32}_{15}P$

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

However, when we talk about the atomic number of the ion, it is not equal to the number of electrons as electron can be gained or loosed.

This is why, more appropriately, the number of the protons which are present in the nucleus of the atom is called the atomic number.

Thus, atomic number of phosphorus = 15

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Given, Mass number = 32

Thus, the symbol of the isotope is:-

$^{32}_{15}P$

8 0

3 years ago