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Shalnov [3]
3 years ago
12

A chemist adds of a mercury(I) chloride solution to a reaction flask. Calculate the micromoles of mercury(I) chloride the chemis

t has added to the flask.
Chemistry
1 answer:
Georgia [21]3 years ago
7 0

Answer:

3.383x10⁻³ micromoles of HgCl

Explanation:

<em>The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.</em>

<em />

The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:

0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.

Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:

3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =

<h3>3.383x10⁻³ micromoles of HgCl</h3>
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In the lab you weigh out 76.02 g of Iron (Fe). How many moles of Iron do you have in the sample. (Your answer must have a unit..
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Answer:

We have 1.361 moles in the sample

Explanation:

Mass of iron = 76.02g

Molar mass of iron = 55.845 g/ mole ( This we can find in the periodic table, and menas that 1 mole of iron has a mass of 55.845 g).

To calculate the number of moles we will use following formula:

moles (n) = mass / molar mass

moles iron = 76.02g / 55.845 g/ mole

moles iron = 1.36127 moles

To use the correct number of significant digits we use the following rule for multiplication and division :

⇒ the number with the least number of significant figures decides the number of significant digits.

⇒76.02 has 4 digits ( 2 after the comma) and 55.845 has 5 digits (3 after the comma).

⇒ this means 1.361 moles

We have 1.361 moles in the sample

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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

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