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Travka [436]
2 years ago
14

17.42 mL of 0.1M NaOH was needed to reach the endpoint in the titration. How many moles of NaOH are in 17.42 mL of 0.1M NaOH

Chemistry
1 answer:
Varvara68 [4.7K]2 years ago
8 0

Answer:

0.001742moles

Explanation:

1000ml of NaOH contain 0.1moles

17.42ml of NaOH contain (0.1*17.42)/1000 moles

= 0.001742moles

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A scientist has two samples of gas: The first sample contains one mole of argon atoms and has a mass of 39.948 g; the second sam
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There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.

Explanation:

39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.

4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.

We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.

So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.

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How many moles are in 39.5 grams of Lithium?
Blizzard [7]

Answer:

185.05 g.

Explanation

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2LiCl → 2Li + Cl₂

It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.

We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.

n = (30.3 g) / (6.941 g/mole) = 4.365 moles.

Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.

Using cross multiplication:

2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.

??? moles of LiCl → 4.365  moles of Li.

The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.

Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).

Molar mass of LiCl = 42.394 g/mole.

mass = n x molar mass = (4.365 x 42.394) = 185.05 g.

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