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pshichka [43]
3 years ago
9

A certain weak acid, ha, has a ka value of 3.6×10−7. part a calculate the percent ionization of ha in a 0.10 m solution. express

your answer to two significant figures and include the appropriate units. view available hint(s)
Chemistry
1 answer:
slega [8]3 years ago
8 0
Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
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<h2><em>no</em></h2>

Explanation:

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7 0
3 years ago
Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283
Lyrx [107]
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
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- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
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6 0
3 years ago
Question 2 (1 point)
gregori [183]

Answer:

John Dalton

Explanation:

Both John Dalton and Democritus thought that the atom was an indivisible sphere until J.J. Thompson came out with the plum pudding model. Hope I helped!

5 0
3 years ago
What do a mole of magnesium (Mg) and a mole of iron (Fe) have in common?
azamat
What they have in common is that they both have the same number of atoms.
9 0
2 years ago
Type the correct answer in the box. Express your answer to three significant figures. Iron(II) chloride and sodium carbonate rea
Ad libitum [116K]

Answer:

The reaction can produce 287 grams of iron(II) carbonate

Explanation:

To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-

<em>Moles FeCl2:</em>

1.24L * (2.00mol / L) = 2.48 moles FeCl2

As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles

<em>Mass FeCO3:</em>

2.48mol * (115.854g / mol) =

<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>
5 0
2 years ago
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