Answer:
2.60 g of H₂ and 20.8 g of O₂ are produced in the decomposition of 23.44 g of water
Explanation:
Water decomposition is:
2H₂O → 2H₂ + O₂
We convert the mass of water, to moles:
23.44 g . 1 mol/18 g = 1.30 moles
Ratio is 2:2 with hydrogen and 2:1 with oxygen. Let's make rules of three:
2 moles of water can produce 2 moles of hydrogen gas and oxygen gas
Then, 1.30 moles will produce:
(1.30 . 2) /2 = 1.30 moles of H₂
(1.30 . 1) /2 = 0.65 moles of O₂
We convert the moles to mass
1.30 moles of H₂ . 2g / 1mol = 2.60 g of H₂
0.65 moles of O₂ . 32 g / 1 mol = 20.8 g of O₂
Answer:
Thats the solution to the question
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
