How many grams would be in 4.45 x 1023 molecules of N203

56.24g

Explanation:

To find the mass of N2O3 in 4.45 x 10^23 molecules, it must first be converted to moles by dividing the number of molecules in N2O3 by Avagadro's number (6.02 × 10^23).

number of moles in N2O3 = 4.45 x 10^23 ÷ 6.02 × 10^23

n = 4.45/6.02 × 10^(23 - 23)

n = 0.74 × 10^0

n = 0.74moles.

Using the formula below to find the mass of N2O3;

mole = mass ÷ molar mass

Molar mass of N2O3 = 14(2) + 16(3)

= 28 + 48

= 76g/mol

mass = mole × molar mass

Mass = 0.74 × 76

Mass = 56.24g

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.