Given:
0.607 mol of the weak acid
0.609 naa
2.00 liters of solution
The solution for finding the ph of a buffer:
[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25
pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.
Because the cord is wrapped in rubber or plastic and they are not good conductors
Answer:
A) M = 100X
B) M = 36X
C) M = 178.88X
Explanation:
Given data:
ASTM grain size number 7
a) total grain per inch^2 - 64 grain/inch^2
we know that number of grain per square inch is given as
where M is magnification, n is grain size
therefore we have
solving for M we get
M = 100 X
B) total grain per inch^2 = 500 grain/inch^2
we know that number of grain per square inch is given as
where M is magnification, n is grain size
therefore we have
solving for M we get
M = 36 X
C) Total grain per inch^2 = 20 grain/inch^2
we know that number of grain per square inch is given as
where M is magnification, n is grain size
therefore we have
solving for M we get
M = 178.88 X
Answer:
1) Write the balanced equation:
2C2H6 + 7O2 ---> 4CO2 + 6H2O
2) Determine limiting reagent:
C2H6 ⇒ 13.8 g / 30.0694 g/mol = 0.45894 mol
O2 ⇒ 45.8 g / 31.9988 g/mol = 1.4313 mol
C2H6 ⇒ 0.45894 / 2 = 0.22947
O2 ⇒ 1.4313 / 7 = 0.20447
Oxygen is limiting.
3) Determine theoretical yield of water:
The oxygen : water molar ratio is 7 : 6
7 is to 6 as 1.4313 mol is to x
x = 1.2268286 mol of water
4) Convert moles of water to grams:
1.2268286 mol times 18.015 g/mol = 22.1 g (to three sig figs)
Solution to (b):
14.2 g / 22.1 g = 64.2%
Explanation:
The milk has a mass concentration of 17 g/l of fat.