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2 years ago

Doy CORONITA a quien me ayude rapido que es un examen porfaaa

Chemistry

1 answer:

rodikova [14]2 years ago

8 0

Answer:

thank you for the points

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What are the three types of symbiotic relationships?

krok68 [10]

Answer: mutualism, commensalism, and parasitism

Explanation: with mutualism, both partners benefit. With commensalism, only one species benefits while the other is neither helped nor harmed. With parasitism, one organism (the parasite) gains benefits, while the other (host) suffers.

6 0

2 years ago

Read 2 more answers

HELP ME!! THANH YOU If an atom has 80 protons and 121 neutrons, and is not in an excited state, how many electrons would Bohr sa

wariber [46]

I think it might be 80

5 0

3 years ago

I didn’t study for this lol if your answer correct I’ll mark ur answer brainliest

kotykmax [81]

Answer:

Explanation:

3 is the answer is did this a little while ago

4 0

3 years ago

If the total population of Asian women was 5000in the 2008 how many females suffered a stroke?

Finger [1]

Answer:

Hi there!

I believe you are missing an attachment to this question however

I strongly believe that the answer you are looking for is 85.

Explanation:

If you provide the graph, the rate of strokes in Asian women will be 17 per 1,000 women so all you have to do is multiply 17 by 5 and you get 85

4 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