Answer:
There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
Explanation:
Molarity of the solution = 2.20 M
Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
Hello!
Ok so for this problem we use the ideal gas law of PV=nRT and I take it that the scientist needs to store 0.400 moles of gas and not miles.
So if we have
n=0.400mol
V=0.200L
T= 23degC= 273k+23c=296k
R=ideal gas constant= 0.0821 L*atm/mol*k
So now we rearrange equation for pressure(P)
P=nRT/V
P=((0.400mol)*(0.0821 L*atm/mol*k)*(296k))/(0.200L) = 48.6 atm of pressure
Hope this helps you understand the concept and how to solve yourself in the future!! Any questions, please feel free to ask!! Thank you kindly!!!
B. An equal amount of protons and electrons
The correct response is C. Chlorine forms 1 covalent bond while Oxygen forms 2 covalent bonds.
Answer
solubility product = 3.18x 10^-7
Explanation:
We were given the pressure in torr then we need to convert to atm for consistency, ten we have
21torr/760= 0.0276315789 atm
21 Torr = .0276315789 atm
P = i M S T
M = P / iRT
Where p is osmotic pressure
T= temperature= 25C+ 273= 298K
for XY vanthoff factor i = 2
S = 0.0821 L-atm / mol K
M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)
M = 0.000564698046 mol/liters
solubility= 0.000564698046 mol/liters
Ksp = [X+][Y-]
Ksp = X^2
Ksp = [Sr^+2] * [SO4^-2]
Ksp = X^2
Ksp = (0.000564698046)^2
Ksp = 3.18883883 × 10-7
Ksp = 3.18x 10^-7
solubility product = 3.18x 10^-7
Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7
