Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
<span>4 I</span>₂<span>+ 9 O</span>₂<span> = 2 I</span>₄<span>O</span>₉
Reactants :
I₂ , O₂
Products :
I₄O₉
hope this helps!
1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
A stock solution is the most concentrated
Answer:
First
divide each element by its Molecular Mass to get their respective moles
Then Divide through by the lowest of the moles
You'll have the ratio of Carbon Hydrogen and Oxygen to be
C2H3O
Given Molecular Mass=184.27
C2H3On=184.27
n(12x2 + 1x3 + 16) =184.27
Evaluating this... You'll have n=4.3
Pls check if you assigned the correct value to each element
