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gizmo_the_mogwai [7]
3 years ago
14

Calculate the number of molecules in 0.75 moles of CH₄. x 10²³ molecules

Chemistry
1 answer:
spin [16.1K]3 years ago
7 0

<u>Ans :</u>4.5 * 10²³ molecules

<u>Given:</u>

Moles of CH4 = 0.75

<u>To determine;</u>

The number of molecules of CH4 in 0.75 moles

<u>Explanation:</u>

1 mole of CH4 contains Avogadro's number i.e. 6.023*10²³ molecules

Therefore, 0.75 moles would correspond to :

= 0.75 moles * 6.023*10²³ molecules/1 mole

= 4.5 * 10²³ molecules

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Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o
vivado [14]

<u>Answer:</u>

<u>For 1:</u> The amount of potassium iodate that were titrated is 2.5\times 10^{-4} moles

<u>For 2:</u> The amount of sodium thiosulfate required is 1.25\times 10^{-4} moles

<u>Explanation:</u>

  • <u>For 1:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of KIO_3 solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol

Hence, the amount of potassium iodate that were titrated is 2.5\times 10^{-4} moles

  • <u>For 2:</u>

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = \frac{1}{2}\times 0.00025=0.000125mol of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is 1.25\times 10^{-4} moles

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How many grams oxygen are present in 76.8g of CO2
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
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Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

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It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

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Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

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Nucleic acids are made of which of the following?
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