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3 years ago

7

True or False: Passive transport occurs when molecules move from an area of low concentration to high concentration.

1 answer:

SpyIntel [72]3 years ago

4 0

Answer:
False, Bc my sister said it was. Don’t worry she smart

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Which statement best describes London dispersion forces?

sineoko [7]

<span>London dispersion forces is the weakest intermolecular force. It is a temporary force that happens when electrons of two adjacent atoms occupy positions that make atoms form dipoles which are temporary dipoles. This is also referred as dipole-dipole attraction.</span>

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3 years ago

What does B represent?

Mashutka [201]

Answer: Botanical Acid

Explanation:

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3 years ago

Write the name and molecular formula of an organic compounds having its name suffixed with -ol and having 2 carbon atoms in the

777dan777 [17]

Ethanol C₂H₆O

Explanation:

When ethanol (CH₃-CH₂-OH) is heated in the presence of the sulphuric acid (H₂SO₄) it will produce ethylene (CH₂=CH₂ ) and water (H₂O).

CH₃-CH₂-OH → CH₂=CH₂ + H₂O

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sulphuric acid

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4 0

3 years ago

A plate moves 200 m in 10,000 years. What is its rate in cm/year?

Varvara68 [4.7K]

Answer:

The answer is

<h2>2 cm/year</h2>

Explanation:

To find the rate in cm/year we must first convert 200 m into cm

1 m = 100 cm

if 1 m = 100 cm

Then 200 m = 200 × 100 = 20 ,000 cm

So the rate is

<h2> $\frac{20000}{10000}$ </h2>

<u>Reduce the fraction with 10,000</u>

We have the final answer as

<h3>2 cm/year</h3>

Hope this helps you

3 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago