They will stay the same through the laws of conservation of mass even when they have changed state so they will have the reactant of Ice will be the same amount as the product of water
So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
Answer:
4.5 g/L.
Explanation:
- To solve this problem, we must mention Henry's law.
- Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
- It can be expressed as: P = KS,
P is the partial pressure of the gas above the solution.
K is the Henry's law constant,
S is the solubility of the gas.
- At two different pressures, we have two different solubilities of the gas.
<em>∴ P₁S₂ = P₂S₁.</em>
P₁ = 525.0 kPa & S₁ = 10.5 g/L.
P₂ = 225.0 kPa & S₂ = ??? g/L.
∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.
Answer:
0.5 M
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 80 g
Volume of solution = 4 L
Molarity =?
Next, we shall determine the number of mole in 80 g of NaOH. This can be obtained as follow:
Mass of NaOH = 80 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 80 / 40
Mole of NaOH = 2 moles
Finally, we shall determine the molarity of the solution. This can be obtained as follow:
Mole of NaOH = 2 moles
Volume of solution = 4 L
Molarity =?
Molarity = mole / Volume
Molarity = 2/4
Molarity = 0.5 M
Therefore, the molarity of the solution is 0.5 M.
