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
If the diameter is 9 cm the radius is 4.5 cm (half the diameter). The area of a circle is given by the formula: 4πr^2, where π is the constant ratio of circumference to diameter (3.14159...).
3.14 cm x 4.5 cm = 14.13 cm^2. Remember, the units must be squared to represent area. 14.13 is extended to the nearest hundredth.
Answer:
huh'?
Step-by-step explanation:
Hello from MrBillDoesMath!
Answer: (4b-6) log(a) = log (3c + d)
Discussion:
Take the log of both sides of the equation:
log ( a ^(4b-6)) = log (3c + d)
As the log functions causes exponents to become multiplexers, this equation is the same as
(4b-6) log(a) = log (3c + d)
Thank you,
MrB
Answer:
Volume of cylinder = 
Step-by-step explanation:
Given : A cylinder fits inside a square prism.
To find : The volume of cylinder
Solution : Refer the attached graph.
Area of circle = 
Area of square = 
Side of square = diameter of circle= 
Diameter = 2r
∴ Area of square= 
Area of circle is 
of area of square.
Volume is always = area × height
Volume of prism = Area of square × h = 
Volume of cylinder = Area of circle × h =
Now, rate
⇒Volume of cylinder is of Volume of prism.
Volume of Cylinder =
Volume of cylinder = 
Volume of cylinder =
