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djverab [1.8K]
3 years ago
13

A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

s solution?
Chemistry
1 answer:
Naddika [18.5K]3 years ago
5 0
Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span>molality =n(solute) / m(solvent) =\ \textgreater \  4.8m = 4.8mole / 1kg
= 4.8 mole * 98 g/mole = 470g
m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \=\ \textgreater \  Molarity = 4.8 mole / 1.2 L = 4 M m(H2SO4)  which is =<span>470g

</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
Molarity = n(solute) / V(solution) =\ \textgreater \
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Find the formula for the compound that contains 72.40% iron and 27.60% oxygen.
Vikki [24]
Oxygen = 16
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3 years ago
Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g). Use the following information: : +279.9 kJ
Len [333]

Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

5 0
3 years ago
2. Nitric oxide contains 46.66% nitrogen and 53.34% oxygen. Water contains 11.21% hydrogen and 88.79% oxygen. Ammonia contains 1
Mrrafil [7]

Answer:

The law of reciprocal proportions states that if two elements react individually with a given weight of a third element, the ratio of the masses with which they combine with the third element are either the same or a simple multiple of the ratio of the masses with which they combine with each other

The compounds formed includes;

1) Nitric oxide, NO

Nitrogen = 46.66% × 30.01 = 14

Oxygen = 53.34% × 30.01 = 16

2) Water, H₂O

Hydrogen = 11.21% × 18.01528 = 2

Oxygen = 88.79% × 18.01528 ≈ 16

3) Ammonia, NH₃

Hydrogen = 17.78% × 17.031 ≈ 3

Nitrogen = 82.22% × 17.031 ≈ 14

The ratio of nitrogen to oxygen in nitric oxide = 14:16 = 7:8

The ratio of nitrogen to hydrogen in ammonia = 14:3

The ratio in which hydrogen and oxygen combine with nitrogen = 3/16

The ratio of hydrogen and oxygen combine with each other in water = 2/16

Therefore, the ratio with which hydrogen and oxygen combine with nitrogen, is (2/3) times the ratio with which they combine with each other, which verifies the law of reciprocal proportions

Explanation:

4 0
3 years ago
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