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3 years ago

What is the overall effect of adding a solute to a Solution

Chemistry

1 answer:

user100 [1]3 years ago

8 0

Answer:

By adding the solute in to solution boiling point is increased while freezing point is decreased.

Explanation:

When solute in added into the solvent the boiling point of solvent increases for example,

Water is boiled at 100 °C, when sodium chloride is added its boiling point increased. Ions of salt interact with solvent and prevent the water molecules to escape from the surface and form gas molecules. In order to make it boiled solution must be heated above 100 °C.

But there is different case with freezing point. Freezing point is the state in which substance converted into the solid. At given temperature when solute is added into the solvent it prevent the formation of solid. It required time to decrease the temperature first and as the temperature is decreases solid is formed.

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3 years ago

I need help solving this!

zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol$

The given reaction equation is as follows.

$C + 2H_{2} \rightarrow CH_{4}$

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

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Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

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3 years ago

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3 years ago

Read 2 more answers

What volume of 0.500 m h2so4 is needed to react completely with 20.0 ml of 0.400 m lioh?

ExtremeBDS [4]

The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
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