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2 years ago

Below is a word equation. Please follow directions for each part below.

Chemistry

1 answer:

Airida [17]2 years ago

7 0

Answer:

C2H6 + 02 ---> CO2 + H20

balanced equation -

C2H6 + 7/202 ---> 2CO2 + 3H20

Explanation:

for equation to be balanced, we must have the same amount of each element on both sides. hence the numbers to balance the equation are 7/2, 2 and 3 respectively.

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Express 0.5 mg/mL in ug/uL

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0.5 mg/mL would be expressed as 500 ug/uL

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3 years ago

What is the pH of a solution with [H ] = 1.25´10–10?

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3 years ago

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How many moles of aluminum are in 4.35 moles of al2si4o10(OH)2

Gnoma [55]

Answer:

8.7 mol Al

Explanation:

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3 years ago

Carbon dioxide is treated with coke.

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It reacts with it and creates carbon monoxide and the affects of it weakens

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3 years ago

Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu

borishaifa [10]

Answer:

$\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}$

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹: 2.7 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 2.7 - x x x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}$

2. Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}$

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

2.7 x 0.0441

$K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}$

6 0

3 years ago