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3 years ago

For the balanced equation shown below, how many moles of O2 b

Chemistry

1 answer:

Bas_tet [7]3 years ago

3 0

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Look at the above table.

MariettaO [177]

Answer:

1. A. 0

2. B. 7

3. C. 4

Explanation:

1. charge is equal to the number of protons minus the number of electrons!

2. neutrons is equal to mass number minus atomic number!

3. valence electrons equal 4!

Hope this helped you! :)

7 0

2 years ago

A good hypothesis must be which of the following?

yanalaym [24]

Testable, that is always required for a hypothesis. Otherwise, how can you prove it?

8 0

3 years ago

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7. This part of the brain controls memory, and long term abuse of marijuana can permanently

eduard

Answer:

Hippocampus

Explanation:

The main parts of the brain involved with memory are the amygdala, the hippocampus, the cerebellum, and the prefrontal cortex.

Memory impairment from marijuana use occurs because THC alters how the hippocampus, a brain area responsible for memory formation, processes information.

7 0

2 years ago

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How many atoms of fluorine are present in a molecule of carbon tetrafluoride, cf4?

Ivanshal [37]

There are four atoms of fluorine present

3 0

2 years ago

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ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion

vfiekz [6]

Answer:

Here's what I get

Explanation:

1. Write the chemical equation

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵

Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

$K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}$

3. Set up an ICE table

A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹: 0.35 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.35-x x x

4. Solve for x

$\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}$

Check for negligibility,

$\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}$

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

pH = 9.12

Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

3 0

3 years ago