What is the mass of frutose also known as fruit (c6h12o6) in a 1.40×10^2ml sample of glucose solution that has a concentration of a 1.23m?

m = 129,96 g

Explanation:

c = n / V

n = c * V

n = 1,23 mol/l * 0,14 l

n = 0,722 mol

n = m / M

m = n * M

m = 0,722 mol * 180 g/mol

m = 129,96 g

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 :

what is the question?