a laboratory procedure calls for making 510.0 mL of a 1.6 M KNO3 solution. How much KNO3 in grams is needed
1 answer:
Answer:
82.416 g of KNO ₃ is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.
Explanation:
Since molarity is the number of moles of solute that are dissolved in a given volume, calculated by dividing the moles of solute by the volume of the solution, the following rule of three can be applied: if in 1 L (1,000 mL) of KNO₃ there are 1.6 moles of the compound present, in 510 mL how many moles will there be?
moles= 0.816
Being the molar mass of the elements:
- K: 39 g/mole
- N: 14 g/mole
- O: 16 g/mole
So the molar mass of the compound KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
Now I can apply the following rule of three: if in 1 mole of KNO₃ there are 101 g, in 0.816 moles how much mass is there?
mass= 82.416 grams
<u><em>82.416 g of KNO ₃ is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.</em></u>
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Answer:
(a) 133.33nm
(b) 600nm
(c) 7,600nm
Explanation:
The concentration of Y can be determined by using the formula:
where;
[L] = concentration of the binding ligand.
kd = 400 nm
Thus:
When Y = 0.25; we get :
0.25 (400 + [L]) = [L]
100 + 0.25[L] = [L]
100 = [L] - 0.25 [L]
100 = 0.75 [L]
[L] = 100/0.75
[L] = 133.33 nm
At, Y = 0.6
0.6 (400 + [L]) = [L]
240 + 0.6[L] = [L]
240 = [L] - 0.6 [L]
240 = 0.4 [L]
[L] = 240/0.4
[L] = 600 nm
At, Y = 0.95
0.95 (400 + [L]) = [L]
380 + 0.95[L] = [L]
380 = [L] - 0.95 [L]
380 = 0.05 [L]
[L] = 380/0.05
[L] = 7600 nm