Molarity is given as,
Molarity = Moles / Volume of Solution ----- (1)
Also, Moles is given as,
Moles = Mass / M.mass
Substituting value of moles in eq. 1,
Molarity = Mass / M.mass × Volume
Solving for Mass,
Mass = Molarity × M.mass × Volume ---- (2)
Data Given;
Molarity = 2.8 mol.L⁻¹
M.mass = 101.5 g.mol⁻¹
Volume = 1 L (I have assumed it because it is not given)
Putting values in eq. 2,
Mass = 2.8 mol.L⁻¹ × 101.5 g.mol⁻¹ × 1 L
Mass = 284.2 g of CuF₂
Oxygen and glucose and energy. <span />
Answer:
33.33% = 33%
Explanation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
1 mole of MCO3 will produce → 1 mole of CO2
We need to get the number of mole of CO2:
and when we have 0.22 g of CO2, so number of mole = mass / molar mass
Moles = 0.22 g / 44 g/mol = 0.005 mole
Moles of Mg = moles of CO2 = 0.005 mole
Mass of Mg = moles * molar mass
= 0.005 * 84 /mol = 0.42 g
Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100
=33.33 %
My guess is b for the question
