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Elis [28]
3 years ago
5

What is the total ionic equation for cuso4+2naoh>>cu(oh)2+na2so4

Chemistry
1 answer:
iVinArrow [24]3 years ago
3 0

Answer:

CuSO4 + 2 NaOH = Cu(OH)2 + Na2SO4

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If 50. 0 g of formic acid (HCHO2, ka = 1. 8 x 10^-4) and 30. 0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of sol
Ksju [112]

If 50. 0 g of formic acid (HCHO2, ka = 1. 8 x 10^-4) and 30. 0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, then the pH of this solution is 3.358.

Option C is correct option.

Given,

Given mass of formic acid = 50 g

Given mass of sodium formate = 30 g

Volume of formic acid = 500 ml

Volume of sodium formate = 500 ml

Molar mass of formic acid = 46 g

Molar mass of sodium formate = 68 g

<h3 /><h3>Calculation of concentration of formic acid and sodium formate</h3>

Molar concentration is defined as the ratio of number of moles and the volume of solution.

Concentration of formic acid

Ca = 50/(46×500)

      = 0.00217 m

Concentration of sodium formate

Cb = 30/(68×500)

     = 0.00088 m

Using Henderson Hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

      = 3.75

Substituting the value of pKa we get,

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75-0.392

pH = 3.35

Thus the value of pH of solution containing formic acid and sodium formate is 3.35.

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7 0
2 years ago
¿Por qué se dice que la Revolución Tecnológica de la segunda mitad del siglo XX está relacionada con el Física de la Materia Con
GREYUIT [131]

translate? ill edit this answer once it is

7 0
3 years ago
Why is Cl weaker base
Helen [10]
Because cl atoms are bonded with hydrogen. hydrogen bond is very strong bond
6 0
3 years ago
A 544 mg of a mixture of fluorene and benzoic acid was weighed out and subjected to an extraction and recrystallization. After t
butalik [34]

<u>Answer:</u> The percent composition of fluorene is 33.08 % and benzoic acid is 36.03 %.

<u>Explanation:</u>

We are given:

Mass of mixture = 544 mg

Mass of fluorene after purification = 1.80\times 10^2mg=180mg

Mass of benzoic acid after purification = 196 mg

To calculate the percentage composition of a substance in a mixture, we use the equation:

\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

<u>For fluorene:</u>

\%\text{ composition of fluorene}=\frac{180}{544}\times 100=33.08\%

Hence, the percent composition of fluorene is 33.08 %

<u>For benzoic acid:</u>

\%\text{ composition of benzoic acid}=\frac{196}{544}\times 100=36.03\%

Hence, the percent composition of benzoic acid is 36.03 %

5 0
4 years ago
PLSS HELP ASAP PLSS​
Zepler [3.9K]

molar concentration of Na⁺ ions = 3 M

molar concentration of CO₃⁻ ions = 1.5 M

First the water will solvate the sodium carbonate (Na₂CO₃).

Second the 80 mL of water added will dilute the first solution.

Explanation:

When you dissolve the Na₂CO₃ in water, the following solvation process takes place:

Na₂CO₃ (s) + H₂O (l) → 2 Na⁺ (aq) + CO₃⁻ (aq)

To find the final concentration of Na₂CO₃ solution we use the following formula:

concentration (1) × volume (1) = concentration (2) × volume (2)

concentration (2) = concentration (1) × volume (1) / volume (2)

concentration of Na₂CO₃ solution = 2.5 M × 120 mL / 200 mL

concentration of Na₂CO₃ solution = 1.5 M

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of Na₂CO₃ = 1.5 M × 0.2 L = 0.3 moles

Knowing the solvation process we devise the following reasoning:

if         1 mole of Na₂CO₃ produces 2 moles of Na⁺ ions and 1 mole of CO₃⁻ ions

then   0.3 moles of Na₂CO₃ produces X moles of Na⁺ ions and Y mole of CO₃⁻ ions

X = (0.3 × 2) / 1 = 0.6 moles of  Na⁺ ions

Y = (0.3 × 1) / 1 = 0.3 moles of CO₃⁻  ions

molar concentration = number of moles / volume (L)

molar concentration of Na⁺ ions = 0.6 / 0.2 = 3 M

molar concentration of CO₃⁻ ions = 0.3 / 0.2 = 1.5 M

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molar concentration

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6 0
3 years ago
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