Answer:
x=3
Step-by-step explanation:
There's many ways you can begin this, I prefer getting rid of the fraction first.
Multiply both sides by 3/2 (yes, I switched it on purpose)
You get. . .
5x-6 = 9
To isolate the 5x, add 6 to both sides
5x = 15
Divide both sides by 5 and you get . . .
x=3
Quantitative data is information that can be measured, while qualitative data is descriptive data which can not be measured.
All five measures (the mean, the median, the trimmed mean, the weighted mean and the mode) can be calculated for quantitative data only, because they all use values in order to be calculated.
Answer:
7.1
Step-by-step explanation:
Using the distance formula
d = sqrt ( (x2-x1)^2 + ( y2-y1)^2)
sqrt ( (-5-2)^2 + ( -3- -2)^2)
sqrt( ( -7)^2 + ( -3 +2)^2)
sqrt( ( 49 + 1)
sqrt( 50)
7.071067812
To the nearest tenth
7.1
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