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zhuklara [117]
2 years ago
9

A 0.100 m solution of which one of the following solutes will have the highest vapor pressure? A 0.100 m solution of which one o

f the following solutes will have the highest vapor pressure? KClO4 Ca(ClO4)2 NaCl sucrose Al(ClO4)3
Chemistry
1 answer:
Assoli18 [71]2 years ago
5 0

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

Vapor pressure is inversely proportional to the number of solute particles. Hence, more will be the solute particles lower will be the vapor pressure and vice-versa.

(a)   KClO_{4} \rightarrow K^{+} + ClO^{-}_{4}

It dissociates to give two particles.

(b)  Ca(ClO_{4})_{2} \rightarrow Ca^{2+} + 2ClO^{-}_{4}

Total number of particles it give upon dissociation are 1 + 2 = 3. Hence, it gives 3 particles.

(c)   Al(ClO_{4})_{3} \rightarrow Al^{3+} + 3ClO^{-}_{4}

Total number of particles it give upon dissociation are 1 + 3 = 4. Hence, it gives 4 particles.

(d)  Surcose being a cobvalent compound doe not dissociate into ions. Therefore, there will be only 1 particle is present.

(e)   NaCl \rightarrow Na^{+} + Cl^{-}

Total number of particles it give upon dissociation are 1 + 1 = 2. Hence, it gives 2 particles.

You might be interested in
How many sodium ions are in the initial 50.00-mL solution of Na2CO3
tresset_1 [31]
From other sources, the given mass of the solute that is being dissolved here is 7.15 g Na2CO3 - 10H2O. We use this amount to convert it to moles of Na2CO3 by converting it to moles using the molar mass then relating the ratio of the unhydrated salt with the number of water molecules. And by the dissociation of the unhydrated salt in the solution, we can calculate the moles of Na+ ions that are present in the solution.

Na2CO3 = 2Na+ + CO3^2-

7.15 g Na2CO3 - 10H2O (1 mol / 402.9319 g) (1 mol Na2CO3 / 1 mol Na2CO3 - 10H2O) ( 1 mol Na2CO3 / 1 mol Na2CO3-10H2O ) ( 2 mol Na+ / 1 mol Na2CO3) = 0.04 mol Na+ ions present
8 0
2 years ago
Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and
user100 [1]

Answer:

The answers to the question are

a. 166.64 ° F

b. 217990.08 J/hour or 60.55 J/s = 60.55 watts

c. 13.C

Explanation:

a. To solve the question we list out the given variables thus

mass of grain = 16.5 lbs

Temperature of grain = 67 °F

Volume of hot water = 5 gals = ‪0.02273‬ m³

Equilibrium temperature of the mixture = 154 °F

Specific heat capacity of the grain = 0.44 times specific heat capacity  of water

Therefore we have

Heat supplied by hot water = heat gained by mixture

Density of the water = 997 kg/m³ which gives

Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × ‪(0.02273‬ m³) = 22.66181 kg

Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -‪67.78 °C) = ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -‪19.44)

= 95172 × (Tₓ -‪67.78 °C) =668205.7536 J

(Tₓ -‪67.78 °C) = 7.02 from where Tₓ = 74.80 °C = ‪166.64 ° F

The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F

b. Where the mixture lost two degrees we have

22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2  °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s

c. To find out how much it cost we have

Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have

22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-‪43.33 °C) = 2994745.92 J

Energy lost during the heating = 10% = 299474.59 J

Total energy supplied 2994745.92 J + 299474.59 J  = 3294220.5 J

Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW

Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost

1.168 kW ×47/60×15 = 13.C

therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F

6 0
1 year ago
What type of chemical reaction is this? Cl2(g) + 2KBr(aq) - 2KCl(aq) + Br2(l)
Sindrei [870]
C. Single-replacement

Chlorine replaces Bromine in KBr.
7 0
2 years ago
Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord
Marina CMI [18]

Answer:

M_{acid}=0.563M

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the acid to obtain:

M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M

Regards!

5 0
2 years ago
Explain how your experimental data for Rf values are, or are not, consistent with your predictions of enantiomeric and diastereo
hichkok12 [17]

Answer:

Answer is explained below.

Explanation:

As (+) menthol and (-) menthol are enantiomers whose physical properties are same except optical activity so we can expect they have similar Rf values.

Whereas diastereomers have different physical properties and different Rf values.

For example when the (+) menthol , (-) menthol, isomenthol and neomenthol undergo TLC (thin layer chromatography) the

Rf values of.(+menthol) = .447

Rf (+isomenthol) = .395

Rf (+neomenthol)= .487

Rf (-menthol) = .434

The above data shows that (+) menthol and (-) menthol have almost same Rf values and vary a little i.e 0.447 and 0.437. So we can conclude them as enantiomers

Whereas (+) menthol or (+) neomenthol or (+) isomenthol i.e 0.447 , 0.395 and 0.487 have different Rf values. We can conclude them as diasteromers.

(+) menthol and (-) menthol - enantiomers

(+) menthol and (+) neomenthol- diastereomers

(-) menthol and (+) isomenthol - diastereomers

3 0
2 years ago
Read 2 more answers
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