No - a precipitation will occur though. Potassium nitrate is soluble in water, so the potassium and nitrate ions will remain spectator ions and stay in solution. Lead (II) hydroxide is not soluble, and will precipitate out of solution to form a solid product.
The volume in liters of 576 grams of SO2 gas at STP is calculated as below
calculate the moles of SO2 = mass/molar mass
= 576 g/64 g /mol = 9 moles
At STP 1mole =22.4 L
what 9 mole =? liters
by cross multiplication
= 22.4 L x 9 moles/ 1moles = 201.6 liters
Answer:
10.5g
Explanation:
First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:
Volume = 250mL = 250/1000 = 0.25L
Molarity = 0.5M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.5 x 0.25
Mole = 0.125 mole
Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:
Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol
Number of mole of NaHCO3 = 0.125 mole
Mass of NaHCO3 =?
Mass = number of mole x molar Mass
Mass of NaHCO3 = 0.125 x 84
Mass of NaHCO3 = 10.5g
Therefore, 10.5g of NaHCO3 is needed.
