Answer:
The answer to your question is P = 1.357 atm
Explanation:
Data
Volume = 22.4 L
1 mol
temperature = 100°C
a = 0.211 L² atm
b = 0.0171 L/mol
R = 0.082 atmL/mol°K
Convert temperature to °K
Temperature = 100 + 273
= 373°K
Formula
Substitution
Simplify
(P + 0.0094)(22.3829) = 30.586
Solve for P
P + 0.0094 =
P + 0.0094 = 1.366
P = 1.336 - 0.0094
P = 1.357 atm
Explanation:
The given data is as follows.
= 10 mM =
M
= 750 ml,
= 5 ml
= ?
Therefore, calculate the molarity of given NaCl stock as follows.
= 1.5 M
Thus, we can conclude that molarity of given NaCl stock is 1.5 M.
Answer:
11.3 g.
Explanation:
Hello there!
In this case, since the combustion of butane is:
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
Therefore, the resulting mass of water is:
Best regards!