POH is defined as -log([OH-])
pOH = -log([OH-]) = -log(7.9 x 10^-4) = 3.1
Hope that helps! :)
3.2 g KClO3
Explanation:
1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)
= 0.0032 mol C12H22O11
0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)
= 0.026 mol KClO3
Therefore, the minimum amount of KClO3 needed is
0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)
= 3.2 g KClO3
Answer:
0.012atm pressure
Explanation:
pressure is inversly proportional to volume. Mathematically,
V1/P1=V2/P2
<u>Answer:</u> The correct option is C) 1.68 mol/L
<u>Explanation:</u>
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(1)
Given values:
Given mass of 
= 150 g
Molar mass of = 180 g/mol
Volume of the solution = 0.50 L
Putting values in equation 1, we get:
Hence, the correct option is C) 1.68 mol/L
Hydrogen gas and oxygen gas react to form liquid water according to the following equation:
2H₂ + O₂ → 2H₂O
a. Converting our given masses of each gas to moles, we have:
(25 g H2)/(2 × 1.008 g/mol) = 12.4 mol H2; and
(25 g O2)/(2 × 15.999 g/mol) = 0.781 mol O2.
From the equation, two moles of H2 react with every one mole of O2. To fully react with 12.4 moles of H2, as we have here, one would need 6.2 moles of O2, which is far more than what we're actually given. Thus, the oxygen is our limiting reactant, and as such it will be the first reactant to run out.
b. Since O2 is our limiting reactant, we use it for determining how much product, in this case, H2O, is produced. From the equation, there is a 1:1 molar ratio between O2 and H2O. Thus, the number of moles of H2O produced will be the same as the number of moles of O2 that react: 0.781 moles of H2O. The mass of water produced would be (0.781 mol H2O)(18.015 g/mol) ≈ 14 grams of water (the answer is given to two significant figures).
c. Since the hydrogen reacts with the oxygen in a 2:1 ratio, twice the number of moles of oxygen in hydrogen is consumed: 0.781 mol O2 × 2 = 1.562 mol H2. Since we began with 12.4 moles of H2, the remaining amount of excess H2 would be 12.4 - 1.562 = 10.838 mol H2. The mass of the excess hydrogen reactant would thus be (10.838 mol H2)(2 × 1.008 g/mol) ≈ 22 grams of hydrogen gas (the answer is given to two significant figures).
