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3 years ago

HELP PLEASEE!

Chemistry

2 answers:

Sever21 [200]3 years ago

6 0

Answer:

drought

Explanation:

drought is the dryest of all of the options

Elanso [62]3 years ago

5 0

Answer:

drought

Explanation:

I did the activity

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Atoms that have the same number of outer electrons

BARSIC [14]

I can see two answers, I’d go with D, but all neutral atoms, of the same element would have the same number of outer electrons. However, if you consider that some of the atoms might be ions, that would eliminate B.

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3 years ago

What is the density of a sample of a substance with a volume of 120ml<br> and a mass of 90g?

tigry1 [53]

Answer:

<h3>The answer is 0.75 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 90 g

volume = 120 mL

We have

$density = \frac{90}{120} = \frac{9}{12} = \frac{3}{4} \\$

We have the final answer as

Hope this helps you

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3 years ago

Suppose that a dependent quantity increases when the independent quantity is decreased. This kind of proportional relationship b

Anna71 [15]

I thinks it’s not this article we should cajnges it

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3 years ago

The advantage of asexual reproduction is.

Nataly_w [17]

B. reproduction doesn’t require mate

6 0

3 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

3 years ago