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2 years ago

Which statement accurately describes part of the dissolving process of a polar solute in water?

Chemistry

1 answer:

Nonamiya [84]2 years ago

5 0

Answer: -Water molecules surround solute molecules.

Explanation: The solubility of substances is governed by: Like dissolves like, which states that polar compounds are soluble in polar solvents and non polar compounds are soluble in non polar solvents.

Water being a polar solvent is able to dissolve only polar solutes and not able to dissolve non polar solutes.

The polar solutes when dissolved in water are surrounded by water molecules and due to high dielectric constant of water the polar molecules are separated and thus get dissolved in water.

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A _____ is a new substance formed in a chemical reaction.

Lorico [155]

B. product
Think about it like this: a company produces products, not coefficients.

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3 years ago

The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to conve

coldgirl [10]

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

$\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%$

5 0

4 years ago

Consider the two electron arrangements for neutral atoms A and B. The outer electron of atom B has moved to a higher energy stat

romanna [79]

B I think the answer

6 0

4 years ago

A student prepares a solution of Potassium Nitrate (KNO3) containing 95g at 40 C. This solution is -

Sholpan [36]

Answer:

unsaturated

Explanation:

4 0

2 years ago

What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent

dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

$\text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}$

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, $\text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}$

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

8 0

4 years ago