28Which is true?

A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution

Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions× $\frac{1(NH_{4})_{2}SO_{4}}{3Ions}$ = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄× $\frac{132,14g}{1mol}$ = 34,6g of (NH₄)₂SO₄

I hope it helps!

7 0

3 years ago

You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl

bulgar [2K]

Answer: Mass of $CO_2$ produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)$

Mass or reactants = Mass of $CaCO_3$ + mass of $HCl$ = 16.00 + 64.80 = 80.80 g

Mass of products = mass of aqueous solution + mass of $CO_2$ + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of $CO_2$ produced in this reaction was 6.56 grams

7 0

3 years ago

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago

If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?

Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

3H2 + N2 ------> 2NH3

from reaction 3 mol 1 mol

given 11.8 mol 1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

3H2 + N2 ------> 2NH3

from reaction 1 mol 2 mol

given 1.01 mol x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0

3 years ago

The density of a pure substance is its mass per unit volume. The density of cresol has been measured to be 1024 g/L . Calculate

Ierofanga [76]

Answer: The mass of 405 ml of cresol is 415 grams

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{volume}$

Given : Density of cresol = 1024 g/L

Volume of cresol = 405 ml = 0.405 L ( 1L=1000ml)

Putting in the values we get:

$1024g/L=\frac{mass}{0.405L}$

$mass=1024g/L\times 0.405L=415g$

Thus mass of 405 ml of cresol is 415 grams

3 0

2 years ago