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AlekseyPX
2 years ago
14

HELP really struggling with this one

Chemistry
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

the answer is B no need to struggle ur answer is correct

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Water is a polar molecule, meaning the covalent bonds in it share electrons unequally. They spend more time around the oxygen, making it positive and the hydrogens negative. This helps water stick to surfaces, a property called adhesion. The hydrogens of one molecule are also attracted to the oxygens of another. This is called cohesion, and it helps water form droplets
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3 years ago
Please comment if you are sure about your answer . Thank you
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4 years ago
Why does solid change to liquid when heat is added
Sophie [7]
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4 0
3 years ago
A sample of nitrogen gas, (N2), occupies 45.0 mL at 27.00∘C and 80.0 kPa. What will be the pressure if the gas is cooled to −73.
maxonik [38]

Answer:

The new pressure is 53.3 kPa

Explanation:

This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.

T° increase → Pressure increase

T° decrease → Pressure decrease

In this case, temperature was really decreased. So the pressure must be lower.

P₁ / T₁ = P₂ / T₂

80 kPa / 300K =  P₂/200K

(80 kPa / 300K) . 200 K = P₂ →  53.3 kPa

6 0
3 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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