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3 years ago

What is the concentration of a solution with a volume of 660L that contains 33.4g of AlCO3?

Chemistry

1 answer:

lana [24]3 years ago

7 0

Answer:

The concentration of the solution is 5.8168 × $10^{-4}$ mol. $dm^{-3}$

Explanation:

Here, we want to calculate the concentration of the solution.

The unit of this is mol/dm^3

So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3

The number of moles = mass/molar mass

molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol

Number of moles = 33.4/87 = 0.384 moles

This 0.384 moles is present in 660 L

x moles will be present in 1 dm^3

Recall 1 dm^3 = 1L

x * 660 = 0.384 * 1

x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3

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3 years ago

Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo

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3 years ago

Read 2 more answers

There are ________ unpaired electrons in the Lewis symbol for a nitride ion.

Anna71 [15]

Answer:

There are no unpaired electrons.

Explanation:

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3 years ago

The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?

Aloiza [94]

Answer:

0.0468 g.

Explanation:

The decay of radioactive elements obeys first-order kinetics.

For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

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3 years ago