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n200080 [17]
3 years ago
7

Need help fast please I am timed and giving 47 points.

Chemistry
2 answers:
Leokris [45]3 years ago
6 0

Answer:

C

Explanation:

antiseptic1488 [7]3 years ago
3 0

Answer:

C - 3

Explanation:

This can be actually tested

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How do catalysts increase the rate of a chemical reaction?
bogdanovich [222]
A, They Lower The Activation Energy.
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Read 2 more answers
Stained glass is homogeneous heterogeneous.​
Serjik [45]

Answer:

It is heterogeneous as it is made up of visibly different substances or phases.

6 0
3 years ago
explain the relationship between the rate of effusion of a gas and its molar mass. methane gas (ch4) effuses 3.4 times faster th
Musya8 [376]

The molar mass of the unknown gas is 184.96 g/mol

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

  • Rate of unknown gas (R₁) = R
  • Rate of CH₄ (R₂) = 3.4R
  • Molar mass of CH₄ (M₂) = 16 g/mol
  • Molar mass of unknown gas (M₁) =?

The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

Cross multiply

(1 / 3.4)² × M₁ = 16

Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

M₁ = 184.96 g/mol

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

#SPJ1

3 0
1 year ago
An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with
Kamila [148]

Answer:

The molar mass of the unknown acid is 89 g/mol

Explanation:

<u>Step 1:</u> The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

<u>Step 2:</u> Data given

Mass of the acid = 0.093 grams

volume = 250 mL

titrates with 0.16 M NaOH

adds 6.5 mL NaOH

<u>Step 3: </u>Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

Moles = 0.00104 moles NaOH

<u>Step 4: </u>Calculate moles of the unknown acid:

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

For 0.00104 moles NaOH we have 0.00104 moles of HA

<u>Step 5: </u>Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

3 0
3 years ago
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