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3 years ago

Compute the boiling point of this solution:

Chemistry

1 answer:

ryzh [129]3 years ago

4 0

Answer:

The boiling point of the solution is 101,42ºC

Explanation:

Formula for boiling point elevation is:

ΔT = Kb · molality . i

Water boils at 100ºC, at 1 atm.

NiSO4 ----> Ni2+ + SO4-2 (assume 100% ionization, the value of i, is 2)

Tº boiling point solution - Tº boiling point = Ebulloscopic constant . molality . Van't Hoff factor.

Tº bp solution - 100ºC = 0,512 ºC/molal . molality . 2

Molality means moles of solute in 1kg of solvent (1000g)

Molar mass NiSO4 = 154,75 g/m

Mass/ Molar mass = Moles --> 21,6 g / 154,75 g/m = 0,139 moles

These moles are in 100 g of H2O, so we have to make a rule of three to find, the moles in 1000g

100 g _________0,139 moles

1000 g _______ 1,39 moles (molality)

Tº bp solution - 100ºC = 0,512 ºC/molal . 1,39molal . 2

Tº bp solution = (0,512 ºC/molal . 1,39molal . 2) + 100ºC

Tº bp solution = 101,42ºC

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How do you describe the milk?

Lady bird [3.3K]

Answer:

emulsion

Explanation:

An emulsion is a mixture of two or more liquids that are normally immiscible. Emulsions are part of a more general class of two-phase systems of matter called colloids.

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3 years ago

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

3 years ago

Assuming an electron is a hydrogen atom can undergo any of the transitions below, thus changing its energy. Which transition is

inn [45]

Answer:

n = 3 to n = 5

Explanation:

According to the Bohr's model of the atom, electrons in an atom absorb energy to move from a lower to higher energy level.

We must note that as we progress away from the nucleus, the energy levels of electrons become closer together. The energy difference between successive levels decreases and the wavelength of light associated with such transitions become longer.

Hence,the absorption of light of the longest wavelength corresponds to n = 3 to n = 5

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2 years ago

Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent for the following equation

bazaltina [42]

The element that gains electrons, becomes reduced.

While the one which loses electrons, becomes oxidized.

In this equation,

CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.

By balancing the equation, we will get:

3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O

Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.

Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.

So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .

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2 years ago

Comare and contrast physical property and chemical property

Nimfa-mama [501]

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2 years ago