This is the balanced eq
N2 + 3H2 -> 2NH3
first you need to find mole of N2 by using
mol = mass ÷ molar mass.
mol N2= 20g ÷ (14.01×2)g/mol
=0.7138mol
then look at the coefficient between H2 and NH3.
it is N2:NH3
1:2
0.7138:0.7138×2
0.7138:1.4276 moles
moles of NH3 = 1.4276 moles
Answer:
a. 174 mL
Explanation:
Let's consider the following reaction.
2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)
We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:
0.1550 L × 0.112 mol/L = 0.0174 mol
The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:
2 × 0.0174 mol = 0.0348 mol
The volume of a 0.200 M KI solution that contains 0.0348 moles is:
0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL
The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5
Answer:
Cl-
Explanation:
Neutral (Cl) have 7 electron, but Cl- have 8 electron due to gain of 1 electron
