Answer:
-6x-12
Step-by-step explanation:
2(-3x-6)
Multiply: 2(-3x)+2(-6)
so answer is -6x-12
Given a = 7 and b = 13, this gives us for c:
6 < c < 20, i maybe be wrong
Volume = π×22<span>×6 = </span><span>75.398223686155 feet^3</span>
Given that the line passes through the two points (-3,4) and (2,8)
We need to determine the equation of the line.
<u>Slope:</u>
The slope of the line can be determined using the formula,
Substituting the points (-3,4) and (2,8) in the above formula, we get;
Thus, the slope of the line is
<u>Equation of the line:</u>
The equation of the line can be determined using the formula,
Substituting the point (-3,4) and in the above formula, we have;
Thus, the equation of the line is
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
