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prohojiy [21]
2 years ago
8

Calculate the pH of a solution that is 0.113 M HCI.

Chemistry
1 answer:
Alex Ar [27]2 years ago
6 0

Simplified concentration

\\ \rm\Rrightarrow 1.13\times 10^{-2}M

Now

\\ \rm\Rrightarrow pH=-log[H^+]

\\ \rm\Rrightarrow pH=-log(1.13\times 10^{-2})

\\ \rm\Rrightarrow pH=-log1.13-log10^{-2}

\\ \rm\Rrightarrow pH=-0.05+2

\\ \rm\Rrightarrow pH=1.95

Highly acidic

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Cora, an electrician, wraps a copper wire with a thick plastic coating. What is she most likely trying to do?.
Vinil7 [7]

Answer:

SHE IS INSULATING THE WIRE SO THAT NO ONE GETS SHOCKED IF THEY ACCIDENTALLY TOUCH IT

7 0
1 year ago
450g of chromium (III) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (AN
Oduvanchick [21]

Answer:

599.26 grams of potassium sulfate will be produced.

Explanation:

Cr_2(SO_4)_3(aq)+2K_3PO_4(aq)\rightarrow 2CrPO_4(s)+3K_2SO_4(aq)

Moles of chromium (III) sulfate = \frac{450 g}{392 g/mol}=1.1480 mol

According to reaction, 1 mole of chromium (III) sulfate gives 3 moles of potassium sulfate.

Then 1.1480 moles of chromium (III) sulfate will give:

\frac{3}{1}\times 1.1480 mol=3.4440 mol

Mass of 3.4440 moles of potassium sulfate:

= 3.4440 mol × 174 g/mol = 599.26 g

599.26 grams of potassium sulfate will be produced.

8 0
3 years ago
In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not
frutty [35]

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

  • A + B  → C

the rate law can be expressed as:  

  • Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁     and     [A]₂ = 2 [A]₁

  • Rate₁ = k[A]₁ᵃ[B]ᵇ    → eq. 1
  • Rate₂ = k[A]₂ᵃ[B]ᵇ    → eq. 2

Dividing eq. 2 by eq. 1 one can get

  • (Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

  • Rate₂ = 2 *Rate₁     and     [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

  • (2) = (2)ᵃ
  • taking log of both sides
  • log (2) = a Log (2)
  • 0.693 = a * 0.693
  • a =1  

∴ order of reaction with respect to A is first (=1)        →     (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁     and     [B]₂ = 2 [B]₁

  • Rate₁ = k[A]ᵃ[B]₁ᵇ    → eq. 1
  • Rate₂ = k[A]ᵃ[B]₂ᵇ    → eq. 2

Dividing eq. 2 by eq. 1 one can get

  • (Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

  • Rate₂ = Rate₁     and   [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

  • (1) = (2)ᵇ
  • taking log of both sides
  • log (1) = b Log (2)
  • 0 = 0.693 * b
  • b = 0

∴ order of reaction with respect to B is zero         →     (2)

So, from 1 and 2  the right choice is rate = k[A]¹[B]⁰= k[A]

6 0
3 years ago
How is a mole defined?​
sattari [20]

A mole of a substance is defined as: The mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000g of 12C. Fundamental units change depending on the substance.

6 0
3 years ago
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