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riadik2000 [5.3K]
3 years ago
6

You have 100 mL of a 12M solution of HCl, and you need to dilute it to 1.5M for an experiment. How many liters will your new sol

ution be?
Chemistry
1 answer:
sashaice [31]3 years ago
8 0

Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

M after dilution = 1.5 M, V after dilution = ??? mL.

∵ (MV)before dilution of HCl = (MV)after dilution of HCl

∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)

<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>

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Answer:

Here's what I get  

Explanation:

A plant extract is a mixture because it contains different substances: acetone or ethanol, chlorophylls A and B, carotene and xanthophylls.

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You can separate the pigments by paper, thin layer, or column chromatography.

Many schools use paper chromatography, because paper is cheap.

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3 years ago
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heightens

Explanation:

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How many electrons are in the highest occupied energy level of these atoms?
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3 years ago
A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution
Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

<em></em>

I hope it helps!

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