A substance is analyzed and found to consist of 0.360 g of Na, 0.220 g of N, and 0.752 g of O. How many moles of Na, N, and O are present? What is its empirical formula? What is its name?

1. The number of mole of each element present = 1mole of Na, 1mole of N and 3moles of O

2. The empirical formula is NaNO3.

3. The name of the compound is sodium trioxonitrate (V)

Explanation:

Mass of Na = 0.360g

Mass of N = 0.220g

Mass of O = 0.752g

Number of mole present for each elements is given below:

For Na:

Molar Mass of Na = 23g/mol

Number of mole = Mass /Molar Mass

Number of mole of Na = 0.360/23 = 0.0157mol

For N:

Molar Mass of N = 14g/mol

Number of mole = Mass /Molar Mass

Number of mole of N = 0.220/14 = 0.0157mol

For O:

Molar Mass of O = 16g/mol

Number of mole = Mass /Molar Mass

Number of mole of = 0.752/16 = 0.047mol

Next we'll divide the moles of Na, N, and O obtained by the smallest mole among them i.e 0.0157

For Na:

0.0157/0.0157 = 1

For N:

0.0157/0.0157 = 1

For O:

0.047/0.0157 = 3

The number of mole of each element present = 1mole of Na, 1mole of N and 3moles of O

Therefore, the empirical formula is NaNO3.

The name of the compound is sodium trioxonitrate (V)

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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