Answer:
In the same way, a solution with a pH of 5 contains 10-5mol/l of hydrogen ions, a solution with a pH of 6 contains 10-6mol/l of hydrogen ions, while the solution with a pH of 7 contains 10-7mol/l of hydrogen ions.
Explanation:
Answer:
A.
Explanation:
The paper clip allows elictricity to pass, unlike the eraser or paper
Answer:
5 mol.
Explanation:
Equation of the reaction
2SO2 + 2H2O + O2 --> 2H2SO4
By stoichiometry, 2 moles of SO2 reacted with 2 moles of water and 1 mole of O2 to give 2 mole of sulphuric acid.
Number of moles:
5.0 mol SO2
4.0 mol O2
20.0 mol H2O
Calculating the limiting reagent,
5 mol of SO2 * 1 mol of O2/2 mol of SO2
= 2.5 mol of O2(4 mol of O2 is present)
5 mol of SO2 * 2 mol of H2O/2 mol of SO2
= 5 mol of H2O(20 mol of H2O)
SO2 is the limiting reagent.
Therefore, number of moles of H2SO4 = 5 mol of SO2 * 2 mol of H2SO4/2 mol of SO2
= 5 mol of H2SO4.
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
Answer is: <span>the coefficient of phosphoric acid is 12.
</span>Chemical reaction: P₄S₃ + NO₃⁻ + H⁺ → H₃PO₄ + SO₄⁻ + NO.
Reduction half reaction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O /·38
Oxidation half reaction: P₄S₃ + 28H₂O → 4H₃PO₄ + 3SO₄²⁻ + 44H⁺ + 38e⁻ /·3.
38NO₃⁻ + 152H⁺ + 3P₄S₃ + 84H₂O → 38NO + 76H₂O + 12H₃PO₄ + 9SO₄²⁻ + 132H⁺.
Balnced chemical reaction:
3P₄S₃ + 38NO₃⁻ + 20H⁺ + 8H₂O → 12H₃PO₄ + 9SO₄²⁻ + 38NO.
