Answer:
80.1 grams
Explanation:
Find the molar mass of CH3OH first by using the periodic table values.
12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H
=32.042 so that is the molar mass
Now that you have 2.50 moles of CH3OH, you can calculate the mass in g
2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105
32.042g / 1 mol is the same as 32.042 g/mol
Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH
Answer:
The more concentrated acetic acid buffer has a better buffer capacity because requires more moles of acid or base to change the pH than a more diluted acetic acid buffer.
Explanation:
Buffer capacity is defined as the moles of an acid or base that are needed to change the pH of a buffer in 1 unit.
A more concentrated solution of acetic buffer contains more moles of the acid per liter of solution. A solution that contains more moles of the acetic ion or the acetic acid requires more moles of base or acid to change the pH, that means:
The more concentrated acetic acid buffer has a better buffer capacity because requires more moles of acid or base to change the pH than a more diluted acetic acid buffer.
As depth increases, the density of the layers decreases.
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
