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Mnenie [13.5K]
3 years ago
11

A colligative property of a solution is one that depends 1. on the identity of the solute and the concentration of the solution.

2. only on the concentration of the solution. 3. only on the identity of the solvent. 4. on the identity of the solvent and the concentration of the solution. 5. only on the identity of the solute.
Chemistry
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

The correct answer is option 2.

Explanation:

Colligative property is defined as property which depends upon only on the numbers of particles of solute dissolved in definite amount of  solvent, It do not depend on the nature of the solute.

For example : NaCl solution with 0.4 molal will show same colligative properties as a that of the glucose solution with 0.04 molal concentration.

The following are the examples of colligative property:

1. Relative lowering of vapor pressure.

2. Osmotic pressure

3. Elevation in boiling points

4. Depression in freezing point

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Compound element ratio of 1:2:6<br> Ba(NO3)2+Na2SO4
Karo-lina-s [1.5K]

1:2:6? Ba(NO3)2 +Na2SO4 -----> BaSO4 + 2NaNO3

Ba(NO3)2

BaSO4

NaNO 3

Na2SO4

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3 years ago
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As pH increases, what happens to the hydrogen ion concentration?
levacccp [35]

Answer:

According to libretexts the answer would be B. decreases.

Explanation:

If the hydrogen concentration increases, the pH decreases, causing the solution to become more acidic. This happens when an acid is introduced. ... If the hydrogen concentration decreases, the pH increases, resulting in a solution that is less acidic and more basic

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3 years ago
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Two masses exerting a force on each other is an example of What????
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Newtons universal law of gravitation

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8 0
3 years ago
1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour
lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9  

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

5 0
2 years ago
In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0
3 years ago
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