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Oksi-84 [34.3K]
3 years ago
13

Subjected synonyms please

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
3 0

<em>EXACT</em>

<em>EXPOSE</em>

<em>LEVY</em>

<em>WREAK</em>

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g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u
DIA [1.3K]

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg .  0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

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4 years ago
A piece of limestone erodes due to acid rain. This process can be best described as a......
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Slow chemical change

It is a chemical change because the erosion is due to the chemical reaction between the acid and the in the rain and the calcium carbonate.

It is slow due to the concentration of acid is low.
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3 years ago
What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the
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Answer:

[NaOH} = 0.4 M

Explanation:

In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.

(H₂SO₄, is considered strong, but the first deprotonation is weak)

2NaOH  +  H₂SO₄  →  Na₂SO₄  + 2H₂O

As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.

In the equivalence point we know mmoles of base = mmoles of acid

Let's finish the excersise with the formula

25 mL . M NaOH = 28.2 mL  .  0.355M

M NaOH = (28.2 mL  .  0.355M) / 25 mL → 0.400

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Nitrogen

Explanation:

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