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dolphi86 [110]
2 years ago
11

What are four sources of drinking water for Canadians

Chemistry
1 answer:
Sergeu [11.5K]2 years ago
3 0

Answer:Water sources: groundwater. Water sources: lakes. Water sources: rivers. Water sources: snow and ice.

Explanation:

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

6 0
2 years ago
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solmaris [256]
The shape of the H2O molecule is a Bent Triatomic.
It isn't symmetrical.
The H2O molecule is polar.
6 0
3 years ago
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The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density
mash [69]

Answer:

The density of acetic acid at 30°C = 1.0354_g/mL

Explanation:

specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)

Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)

= 0.9956 g/mL × 1.040

= 1.0354_g/mL

Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.

Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure

7 0
3 years ago
HELLLLP ITS 8TH GRADE SCIENCE AND DUE TODAY
insens350 [35]

Answer:G

Explanation:

I would look it up on safari just to be sure tho, hope this helps

6 0
2 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
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