Calculate the concentration of a solution containing 1.8g of sodium carbonate in 862cm3 of water

0.02 M

Explanation:

Given data:

Mass of sodium carbonate = 1.8 g

Volume of water = 862 cm³ (862/1000 = 0.862 L)

Concentration = ?

Solution:

Formula:

Concentration  = number of moles / volume in L

Number of moles of sodium carbonate:

Number of moles = mass/molar mass

Number of moles = 1.8 g/ 106 g/mol

Number of moles = 0.02 mol

Concentration:

c = 0.02 mol / 0.862 L

c = 0.02 M

answer; i believe that the correct answer is (a) potassium(k)

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 :