Answer:
Step-by-step explanation:
8xC-Bdivide by c
Answer:
I am not sure about this question sry but u can try asking a tutor u don't need to use any points
Answer:
6th floor
Step-by-step explanation:
2-3+7
-1+7
6
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)
moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g
Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05
N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
Class A: 6v + 8b = 202
Class B: 12v + 16b = 284
Solve using the elimination method:
since 6v and 12v are perfect for elimination, multiply the class A equation by 2 so that the van variable cancels out:
12v + 16b = 404
12v + 10b = 284
Then subtract the bottom equation from the top:
6b = 120
b = 20
Now you know that each bus can hold 20 students.
Just plug this into one of the original equations to solve for vans:
6v + 8(20) = 202
6v + 160 = 202
6v = 42
v=7
So then you know that each van can hold 7 students.
Check:
12 (7) + 10 (20) = 284
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