Answer : The mass of reactant remain would be, 0.20 grams.
Solution : Given,
Moles of = 0.40 mol
Moles of = 0.15 mol
Molar mass of = 2 g/mole
First we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 2 mole of
So, 0.15 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
The moles of reactant remain = 0.40 - 0.30 = 0.10 mole
Now we have to calculate the mass of reactant remain.
Therefore, the mass of reactant remain would be, 0.20 grams.
Answer:
The volume of a ball that is 24 cm across is :
d). v = four-thirds pi 12 cubed
Explanation :
Ball is a sphere and volume of sphere is given by :
r = radius of ball
r = 12 cm
Volume is :
Answer:
484.925 grams KClO3.
Explanation:
Given 3.50 moles of KCl as the target amount in the problem, used the coefficient of the balanced chemical reaction involved to determine the number of moles of potassium perchlorate needed.
Mass of potassium perchlorate, g = 3.50 moles KCl x [(2 mol KClO3)/(2 mol KCl)] x [(138.55 g KClO3)/(1 mol KClO3)] = 484.925 grams KClO3.
Its Rhodium....In Period 5 of the Periodic Table.
Answer:
[H⁺] = 2.0 × 10⁻¹² M
Explanation:
Given data:
Concentration of KOH = 0.005M
H⁺ = ?
Solution:
pOH = -log[ OH⁻]
pOH = -log[0.005]
pOH = 2.3
pH + pOH = 14
14 - pOH = pH
pH = 14 - 2.3
pH = 11.7
pH = -log[H⁺]
11.7 = -log[H⁺]
[H⁺] = Antilog (-11.7)
[H⁺] = 2.0 × 10⁻¹² M