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Sever21 [200]
3 years ago
15

what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration​

Chemistry
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

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If the seawater carbonate ion (CO3 2-) concentration is 270 µmol/kg, a) what is the approximate rate of calcification, and b) ap
STALIN [3.7K]

Explanation:

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What is the independent variable for this experiment?
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Answer:

the type of drink they consumed each day

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A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2
Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

<u>Step 1: </u>Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

<u>Step 2</u>: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

<u>Step 3:</u> Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

<u>Step 4: </u>Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

<u>Step 5:</u> Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0
3 years ago
Write short note on
WITCHER [35]

I hope it helps you ❤️❤️❤️❤️

4 0
3 years ago
Read 2 more answers
A student used 1.168g of an unknown weak acid and titrated it against 28.75 mL of 0.105M NaOH to reach the equivalence point. Wh
Vaselesa [24]

Answer:

The molar mass of the unknown acid is 386.8 g/mol

Explanation:

Step 1: Data given

Mass of the weak acid = 1.168 grams

volume of NaOH = 28.75 mL = 0.02875 L

Molarity of NaOH = 0.105 M

Since we only know 1 equivalence point, we suppose the acid is monoprotic

Step 2: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.105 M * 0.02875 L

Moles NaOH = 0.00302 moles

We need 0.00302 moles of weak acid to neutralize the NaOH

Step 3: Calculate molar mass of weak acid

Molar mass = mass / moles

Molar mass = 1.168 grams / 0.00302 moles

Molar mass = 386.8 g/mol

The molar mass of the unknown acid is 386.8 g/mol

3 0
3 years ago
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