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Alina [70]
3 years ago
10

What I said the molarity of an hcl solution if 24.0 ml is completely neutralized by 98.0 ml of 0.250m ca(oh)2

Chemistry
1 answer:
ozzi3 years ago
6 0
If molarity is in mol/dm3 
2HCL+Ca(OH)2=CaCl2+2H2O
Moles of ca(oh)2 = 98/1000 x 0.250
                           = 0.0245 moles
moles of hcl = 0.049 As the 1 is to 2 ratio among them
moles = molarity x volume 
0.049 = molarity x 24/1000
molarity=2.042 mol/dm3 
ANSWER IS 2.042 mol/dm3
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Remember that density refers to the "mass per unit volume" of an object.

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3 years ago
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3 0
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7 0
1 year ago
I NEED HELP ASAP! PLEASE BE GENUINE
love history [14]

1. The molar mass of the unknown gas obtained is 0.096 g/mol

2. The pressure of the oxygen gas in the tank is 1.524 atm

<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>1. How to determine the molar mass of the gas </h3>
  • Rate of unknown gas (R₁) = 11.1 mins
  • Rate of H₂ (R₂) = 2.42 mins
  • Molar mass of H₂ (M₂) = 2.02 g/mol
  • Molar mass of unknown gas (M₁) =?

R₁/R₂ = √(M₂/M₁)

11.1 / 2.42 = √(2.02 / M₁)

Square both side

(11.1 / 2.42)² = 2.02 / M₁

Cross multiply

(11.1 / 2.42)² × M₁ = 2.02

Divide both side by (11.1 / 2.42)²

M₁ = 2.02 / (11.1 / 2.42)²

M₁ = 0.096 g/mol

<h3>2. How to determine the pressure of O₂</h3>

From the question given above, the following data were obtained:

  • Volume (V) = 438 L
  • Mass of O₂ = 0.885 kg = 885 g
  • Molar mass of O₂ = 32 g/mol
  • Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
  • Temperature (T) = 21 °C = 21 + 273 = 294 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Pressure (P) =?

The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by V

P = nRT / V

P = (27.65625 × 0.0821 × 294) / 438

P = 1.524 atm

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

Learn more about ideal gas equation:

brainly.com/question/4147359

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