The different types of motion
Answer:
VH2SO4 = 145.3 mL
Explanation:
Mw BaO2 = 169.33 g/mol
⇒ mol BaO2 = 53.5g * ( mol BaO2 / 169.33 g BaO2) = 0.545 mol BaO2
⇒according to the reaction:
mol BaO2 = mol H2SO4 = 0.545 mol
⇒ V H2SO4 = 0.545 mol H2SO4 * ( L H2SO4 / 3.75 mol H2SO4 )
⇒V H2SO4 = 0.1453 L (145.3 mL)
I would say 4 but you can figure it out or if you have a parent near by to
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Enyzmes speed up all chemical reactions
(Just took an AP Bio test with this question on it)
