Answer:
Solution given:
<6+24=180[co- interior angle]
<6=180-24=156°
<6=156°
Nickels are 5 cents, and dimes are 10 cents.
(By luck..) eventually you can have 12 dimes, meaning you have 120 cents, or $1.20 ... 20-12=8.. meaning you used 12 dimes and are missing 8 more coins. which means you can have 8 nickels which is 40 cents. 120+40=160 cents or $1.60
Hope this helped.
The limit does not exist. Why? Because the left hand limit DOES NOT equal the right hand limit. Let’s double check:
We could use -0.000001 to represent the left hand limit. This is less than 0. We plug in 5x - 8
5(-0.000001) - 8
-0.000005 - 8
-8.000005
If we would continue the limit (extend the zeros to infinity), we would get exactly
-8
That is our left hand limit.
Our right hand limit will be represented by 0.000001. This is greater than 0. We plug in abs(-4 - x)
abs(-4 - (0.000001))
abs(-4.000001)
4.000001
If we would continue the limit (extend the zeros to infinity), we would get exactly
4
4 does not equal -8, therefore
The limit does not exist
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
So if we think of a test tube, it looks sort of like a cylinder. This means that its cross-section would be a circle. To find out how many turns a piece of thread would make around the test tube, we need to find the circumference of the test tube, then divide the length of the string by the circumference.
Step 1) Find the circumference
C = pi x diameter
C = 3.14 x 3
C = 9.42
Step 2) Divide the length of the string by the circumference
90.42 / 9.42 = 9.5987
The string would make approximately 9.60 turns around the test tube.
Hope this helps!! :)
