Answer:
50 g of K₂CO₃ are needed
Explanation:
How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?
We analyse data:
500 g is the mass of the solution we want
10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution
Therefore we can solve this, by a rule of three:
In 100 g of solution we have 10 g of K₂CO₃
In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃
Hey there!:
Volume in mL :
1.71 L * 1000 => 1710 mL
Density = 0.921 g/mL
Therefore:
Mass = Density * Volume
Mass = 0.921 * 1710
Mass = 1574.91 g
Nitrous Acid.
Hyponitrous acid: H2N2O2
Nitric acid: HNO3
Pernitric acid: HNO
To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.
The chemical formula for acido perbromico is HBrO4 or perbromic
acid or bromate. It is an inorganic compound and an oxoacid of bromine. When an
H+, the –ate ion is –ic acid: one less O is –ous acid, ttwo less is hypo- -ous
acid and one more is per- -ic acid.