What is the specific latent heat of fusion for a substance that takes 550 kJ to melt 14 kg at 262 K?

Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant.

DIA [1.3K]

Answer:

The correct option is: A. 0.168 M

Explanation:

Chemical reaction involved:

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) → 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O

Given: For MnO₄⁻ solution-

Number of moles: n₁ = 1, Volume: V₁ = 20.2 mL, Concentration: M₁ = 0.0250 M;

For Fe²⁺ solution:

Number of moles: n₂ = 5, Volume: V₂ = 15 mL, Concentration: M₂ = ?M

To find out the concentration of Fe²⁺ solution (M₂), we use the equation:

$\frac{M_{1}\times V_{1}}{n_{1}} = \frac{M_{2}\times V_{2}}{n_{2}}$

$\frac{0.0250 M\times 20.2 mL}{1} = \frac{M_{2}\times 15 mL}{5}$

$0.505 = M_{2}\times 3$

$M_{2} = \frac{0.505}{3} = 0.168M$

Therefore, the concentration or molarity of Fe²⁺ solution: M₂ = 0.168 M

8 0

3 years ago

The hydrogen bonding in water creates ________, a cohesive force that enables one to slightly overfill a glass with water or all

ziro4ka [17]

Answer: -

Surface Tension

Explanation: -

Surface tension is cohesive force created as a result of hydrogen bonding, that enables a liquid drop to have a minimum surface area.

Due to it being cohesive, the water top surface is concave in nature, allowing us to hence slightly overfill a glass with water.

Due to surface tension, the surface of water behaves like a stretched membrane, allowing dense objects like a length wise steel needle to float on water.

Thus, the hydrogen bonding in water creates __surface tension__, a cohesive force that enables one to slightly overfill a glass with water or allows denser objects, such as a lengthwise steel needle, to float on water

8 0

3 years ago

What happens if a cell cannot do cellular respiration?

Dahasolnce [82]

Answer:

the cell enentually dies because it has no way to survive

Explanation:

8 0

3 years ago

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

ΔH = -793,6 kJ

I hope it helps!

5 0

3 years ago

A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its

Over [174]

In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(0.1510 g) * 100
Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0

3 years ago