125 Each half life it divides by 2 the amount
1000/2=500
500/2=250
250/2=125
<u>Given:</u>
Concentration of Ba(OH)2 = 0.348 M
<u>To determine:</u>
pOH of the above solution
<u>Explanation:</u>
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157
Answer:
33 g of H2
Explanation:
N2 + 3 H2 -> 2 NH3
3 mol H2 -> 2 mol NH3
x -> 11 mol NH3
x= (11 mol NH3 * 3 mol H2)/ 2 mol NH3
x= 16.5 mol H2
1 mol H2 -> 2 g
16.5 mol H2 ->x
x= (16.5 mol H2 * 2 g)/ 1 mol H2
x= 33 g