How many moles are in 10.23 g of PO4-3? How many moles are 8.25 x 10^28 molecules of Na2CO3? What is the mass of 6 moles of CH2O? How many formula units are in 6.34 g of NaCl? How many ions are in 0.25 moles of Cu+2? How many grams are in 3.4 x 10^24 molecules of CH4? How many moles are in 10 mL of water (density of water = 1 g/mL)? (sorry for so many questions in one)

1) 0.1077 moles

2) 137043.2 moles

3) 180.186 grams

4) 6.53×10²² molecules or formula units

5) 1.505×10²³ ions

6) 90.626 grams

7) 0.555 moles

Explanation:

1) The number of moles, n = Mass of the substance/(Molar mass of the substance)

The molar weight of PO₄⁻³ is 94.971 g/mol

The number of moles in 10.23 g of PO₄⁻³ is n =10.23/94.971 = 0.1077 moles

The number of moles in 10.23 g of PO₄⁻³ = 0.1077 moles

2) The number of molecules of Na₂CO₃ in one mole of Na₂CO₃ is given by the Avogadro's number, = 6.02 × 10²³ molecules

Therefore, the number of moles in 8.25 × 10²⁸ molecules of Na₂CO₃ is found by determining how many Avogadro's number of molecules are in 8.25 × 10²⁸ molecules of Na₂CO₃ as follows;

The number of moles in 8.25 × 10²⁸ molecules of Na₂CO₃ = (8.25 × 10²⁸)/(6.02 × 10²³) = 137043.2 moles

The number of moles in 8.25 × 10²⁸ molecules of Na₂CO₃ = 137043.2 moles

3) The molar mass of CH₂O = 30.031 g/mol

Therefore, the mass of 6 of CH₂O = 6 moles × 30.031 g/mol = 180.186 grams

The mass of 6 of CH₂O = 180.186 grams

4) The molar mass of NaCl = 58.44 g/mol

The number of moles of NaCl in 6.34 g of NaCl 6.34/58.44 = 0.1085 moles

1 formula unit of NaCl = 1 molecule of NaCl

The number of molecules in one mole of  a substance =   = 6.02 × 10²³

Therefore, the number of molecules in  0.1085 moles of NaCl = 6.02 × 10^(23) × 0.1085 = 6.53×10²² molecules or formula units

5) The number of ions per mole of any substance is give by Avogadro's number, = 6.02 × 10²³ ions,

The number of ions in 0.25 moles of Cu⁺² = 0.25×6.02 × 10²³ ions = 1.505×10²³ ions

The number of ions in 0.25 moles of Cu⁺² =  1.505×10²³ ions

6) The molar mass of CH₄ = 16.04 g/mol

The number of moles, n, in 3.4×10²⁴ = 3.4×10^(24)/(6.02×10^(23)) = 5.65 moles 5.65*16.04

The mass of 5.65 moles of CH₄ = 5.65 moles × 16.04 g/mol = 90.626 grams

The mass of 3.4×10²⁴ molecules of CH₄ = 90.626 grams

7) The density of water = 1 g/mL

The volume of the water 10 mL

The mass of the water = Volume × Density = 10 mL × 1 g/mL= 10 grams

The molar mass of water = 18.015 g/mol

The number of moles of water in 10 g  of water = mass/(Molar mass) = 10/18.015 = 0.555 moles.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

just did the assignment and the answer was #1 +2 n