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2 years ago

A 20.0-milliliter sample of HCl(aq) is completely neutralized by 32.0 milliliters of 0.50 M KOH(aq).

2 answers:

3 0

You can use the formula M1 x V1 = M2 x V2 where M1 is the molarity of the first substance and V1 is the volume of the substance. M2 is the molarity of the 2nd substance and V2 is the volume of the substance

if substance 1 is HCl and 2 is KOH we can set up the following equation

x moles / liter (unknown) x .02 liters = .5 moles/ liter x .032 liters

x moles / liter x .02 liters = .016 moles

x moles / liter = .016/.02 liters

x moles / liter = .8 moles/ liter

You have . 8M HCL

aliya0001 [1]2 years ago

3 0

Answer:

Explanation:

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bagirrra123 [75]

Answer:

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2 years ago

A student runs an experiment in the lab and then uses the data to prepare an Arrhenius plot of the natural log of the rate const

Thepotemich [5.8K]

Answer:

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Explanation:

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2 years ago

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OLga [1]

Answer is: A. Chemical energy to electromagnetic energy and thermal energy.

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2 years ago

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You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak

Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

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2 years ago

a 0.5674 g piece of copper is added to 10.00 ml; of 16 M HNO3, producing 0.8024 g of copper (ii) nitrate. What is the precent yi

uysha [10]

Idk there just making me answer it

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2 years ago

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