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2 years ago

Match the following aqueous solutions with the appropriate letter.

Chemistry

1 answer:

Alla [95]2 years ago

6 0

Answer:

Explanation:

When a salt is dissolved , it increases the boiling point . Increase in boiling point depends upon number of ions . So it is a colligative property .

.19 m AgNO₃ . Each molecule will ionize into two ions . So effective molar concentration is 0.19 x 2 = .38 m

0.17 m CrSO4.Each molecule will ionize into two ions . So effective molar concentration is 0.17 x 2 = .34 m

0.13 m Mn(NO₃)₂. Each molecule will ionize into three ions . So effective molar concentration is 0.13 x 3 = .39 m

0.31 m Sucrose(nonelectrolyte). Molecules will not ionize . So effective molar concentration is 0.31 x 1 = .31 m

Higher the molar concentration , greater the depression in boiling point .

So lowest boiling point is 0.13 m Mn(NO₃)₂.

second highest boiling point is 0.19 m AgNO3.

Third lowest boiling point is 0.17 m CrSO4

Highest boiling point or lowest depression 0.31 m Sucrose.

a . 4

b . 1

c . 2

d . 3

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You are given a 1.00 molal solution of an unknown solute dissolved in water and told that you must determine if the solute is a

aniked [119]

Answer: Yes, the solute is a weak electrolyte

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-(-2.05))^0C=2.05^0C$ = Depression in freezing point

i= vant hoff factor = ?

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality

$\Delta T_f=i\times K_f\times m$

$2.05=i\times 1.86\times 1$

$i=1.10$

As i for non electrolytes is 1, the given compound must be an electrolyte and a weak electrolyte as the value is only slightly greater than 1.

8 0

3 years ago

The standard cell potential (e°) of a voltaic cell constructed using the cell reaction below is 0.76 v: zn (s) + 2h+ (aq) → zn2+

Temka [501]

Cell reaction overall is Zn(s)+2H+(aq)→Zn²+(aq)+H2(g).
The half-reaction of oxidation is Zn(s)→Zn²+(aq)+2E- and E°zn2+/zn=0.76v
Half reaction reduction 2H+(aq)+2∈-→H2(g) and E°H+/H2+=0.00v
Cell potential is E°cell=E°cathode-E°anode
=E°H+/H2e-E°zn2+/zn
=0.00v-(-0.76v)
=0.76v
Nernst equation
Ecell = -0.059W/N log [zn²+]PH2/[Zn][H+]²
Ecell = 0.66v
[zn²+]=1.0M
1=[Zn]
PH2=1atm
[H+]=?
n = number of moles of electrons transfered in the cell=2mol
Ecell=E°cell -0.059W/n log [Zn²+]/[Zn][H+]²
0.66v = 0.76 - 0.059W/2 log 1.0×1.0/1.0×[H+]²
0.059W/2 log 1/[H+]² = 0.76v-0.66v = 0.10v
log 1/[H+]² =0.10v×2/0.059W =3.4
-2log [H+] = 3.4
log [H+] = 1.7
[H+] =10-n
=0.020
=2.0×10-²M
The concenrtation in cathodic compartments is 2.0×10-²M

6 0

3 years ago

What happens to the concentration of a solution as more solvent is added, but the amount of solute remains the same?

elena55 [62]

Answer: CCCCCCCCCCCCCCCCCCCCCc

6 0

3 years ago

Ammonium carbonate decomposes upon heating according to

BaLLatris [955]

The total volume of the gas produced, given the data from the question is 11.84 L

<h3>How to determine the mole of (NH₄)₂CO₃</h3>

Mass of (NH₄)₂CO₃ = 11.83 g
Molar mass of (NH₄)₂CO₃ = 96 g/mol
Mole of (NH₄)₂CO₃ =?

Mole = mass / molar mass

Mole of (NH₄)₂CO₃ = 11.83 / 96

Mole of (NH₄)₂CO₃ = 0.123 mole

<h3>How to determine the volume of NH₃</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 2 moles of NH₃

Therefore,

0.123 mole of (NH₄)₂CO₃ will decompose to produce = 0.123 × 2 = 0.246 mole of NH₃

Thus, we can determine the volume of NH₃ produced by using the ideal gas equation as follow:

Temperature (T) = 22 °C = 22 + 273 = 295 K
Pressure (P) = 1.02 bar
Gas constant (R) = 0.08314 bar.L/Kmol
Number of mole (n) = 0.246 moles
Volume of NH₃ (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.246 × 0.08314 × 295) / 1.02

Volume of NH₃ = 5.92 L

<h3>How to determine the volume of CO₂</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 1 moles of CO₂

Therefore,

0.123 mole of (NH₄)₂CO₃ will also decompose to produce = 0.123 mole of CO₂

Thus, we can determine the volume of CO₂ produced by using the ideal gas equation as follow:

Temperature (T) = 22 °C = 22 + 273 = 295 K
Pressure (P) = 1.02 bar
Gas constant (R) = 0.08314 bar.L/Kmol
Number of mole (n) = 0.123 moles
Volume of CO₂ (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.123 × 0.08314 × 295) / 1.02

Volume of CO₂ = 2.96 L

<h3>How to determine the volume of H₂O</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 1 moles of H₂O

Therefore,

0.123 mole of (NH₄)₂CO₃ will also decompose to produce = 0.123 mole of H₂O

Thus, we can determine the volume of H₂O produced by using the ideal gas equation as follow:

Temperature (T) = 22 °C = 22 + 273 = 295 K
Pressure (P) = 1.02 bar
Gas constant (R) = 0.08314 bar.L/Kmol
Number of mole (n) = 0.123 moles
Volume of H₂O (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.123 × 0.08314 × 295) / 1.02

Volume of H₂O = 2.96 L

<h3>How to determine the total volume of gas produced</h3>

Volume of NH₃ = 5.92 L
Volume of CO₂ = 2.96 L
Volume of H₂O = 2.96 L
Total volume of gas =?

Total volume = 5.92 + 2.96 + 2.96

Total volume of gas = 11.84 L

Learn more about ideal gas equation:

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Learn more about stoichiometry:

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7 0

1 year ago

Three of the primary components of air are

Darya [45]

Answer:

125.681 torr

Explanation:

The formula to calculate the partial pressure of oxygen:

P(O2) = Pcommon - PCO2 - PN2

1 atm = 760 torr

Therefore you need to find the partial pressure of oxygen:

P(O2) = 760 torr -0.285 torr - 634.034 torr =130.013 torr

- Hope that helped!

8 0

3 years ago