Answer:
4) 0.5507 Mol Ar
5) 44.548 Mol AgNO3
6) 2.16107 Mol Li
Explanation:
Divide = grams to mol and mol to molecules
Multiply = molecules to mol and mol to grams.
3) 22 <em>g of Ar</em> 1 mol 22
-------------------- * ---------------------- = ------------ (turn to decimal) = 0.5507 Mol Ar
39.948 <em>g of Ar</em> 39.948
4) AgNO3 = (Ag; 39.948) (N; 14.007) (O^3; 48)
39.948 + 14.007 + 48 = 101.955
7.4 x 10^23 Mc 6.02 x 10^23 Mc
---------------------- * ------------------------- = 44.548 Mol AgNO3
1 mol
5)
15g 1 Mol 15
------- * ------------------- = --------- (turn to decimal) = 2.16107 Mol Li
6.941 g of Li 6.941
Hey there!
Mass = 3 g
Volume = 3 mL
Density = mass / volume
D = 3 / 3
D = 1,0 g/mL
Answer:
23.8 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of potassium iodide is 166.00 g/mol. The moles corresponding to 150 grams are:
150 g × (1 mol/166.00 g) = 0.904 mol
0.904 moles of potassium iodide are contained in an unknown volume of a 0.0380 mol/L potassium iodide solution. The volume is:
0.904 mol × (1 L/0.0380 mol) = 23.8 L
Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!