Which of the following numerical expressions gives the number of particles in 2.0 g of Ne?

Chemistry

1 answer:

mojhsa [17]3 years ago

7 0

Answer:

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

Explanation:

Step 1: Data give

Mass of Ne = 2.0 grams

Molar mass of neon = 20.18 g/mol

Number of Avogadro = 6.0 *10^23 /mol

Step 2: Calculate number of moles of neon

Moles Ne = Mass of ne / Molar mass of ne

Moles Ne = 2.0 / 20.18 g/mol

Moles Ne = 0.099 moles

Step 3: Calculate nulber of particles

Number of particles = 6.022*10^23 / mol * 0.099 moles = 5.96 *10^22

Number of particles = 6.022*10^23 * (2.0g/ 20.18g/mol)

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

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copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation

a_sh-v [17]

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

$KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2$

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it: