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1 year ago

Which member of each pair is more soluble in water? Why? (a) CH₃CH₂OCH₂CH₃(l) or CH₃CH₂OCH₂(g)

Chemistry

1 answer:

Anna [14]1 year ago

7 0

CH₃CH₂OCH₂ is more soluble in water because it has shorter hydrocarbon chain.

<h3>What is hydrocarbon?</h3>

Hydrocarbon is defined as the compound which contain hydrocarbon and carbon atoms.

The carbon atom attached to each other to form framework and hydrogen atom attach to them in different ways to give different configuration. One of the most popular hydrocarbon compound is diamond.

<h3>Solubility of hydrocarbon in water</h3>

Hydrocarbon is non polar compound whereas water is polar compound. So, hydrocarbon is in soluble in water. But as they have weak intermolecular interactions known as London dispersion forces i.e. Instantaneous dipole-induced dipole interactions.

make them less soluble in water.

Greater the hydrocarbon chain lesser will be the solubility of ketone in water. On the other hand, lesser the hydrocarbon chain greater will be the solubility of ketone in water.

Thus, we concluded that the CH₃CH₂OCH₂ is more soluble in water because it has shorter hydrocarbon chain.

learn more about hydrocarbon:

brainly.com/question/16020705

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2 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

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3 years ago

How many particles would be found in a 89.2 g sample of Na?

julsineya [31]

One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert 89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles

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2 years ago