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Mkey [24]
3 years ago
15

1.4g of calcium oxide obtained by heating limestone was found to contain 0.4g of oxygen.another sample of 3.5g of calcium oxide

obtained by direct combination of calcium and oxygen was found to contain 2.5g of calcium.show that data are in accordance with the law of definite proportion​
Chemistry
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

Solution

1.4 gm of CaO⟶n0.4 gm of Oxygen

1.0 gm of O2⟶3.5 gm of Ca

According to the law of definite proportions

n/m=y/x

⇒1.4/0.4=3.5.

⇒3.5=3.5

Explanation:

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What is the empirical formula for a substance that contains 3.730% hydrogen, 44.44% carbon, and 51.83% nitrogen by mass?\?
Vlada [557]

Given data:

Hydrogen (H) = 3.730 % by mass

Carbon (C) = 44.44%

Nitrogen (N) = 51.83 %

This means that if  the sample weighs 100 g then:

Mass of H = 3.730 g

Mass of C = 44.44 g

Mass of N = 51.83 g

Now, calculate the # moles of each element:

# moles of H = 3.730 g/ 1 g.mole-1 = 3.730 moles

# moles of C = 44.44/12 = 3.703 moles

# moles of N = 51.83/14 = 3.702 moles

Divide by the lowest # moles:

H = 3.730/3.702  = 1

C = 3.703/3.702 = 1

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
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Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
3 years ago
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