Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
The balanced equation for the reaction between Mg and O₂ is as follows
2Mg + O₂ --> 2MgO
stoichiometry between Mg and O₂ is 2:1
number of Mg reacted - 4.00 mol
if 2 mol of Mg reacts with 1 mol of O₂
then 4.00 mol of Mg requires - 1/2 x 4.00 = 2.00 mol of O₂
then the mass of O₂ required - 2.00 mol x 32.0 g/mol = 64.0 g
64.0 g of O₂ is required for the reaction
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃] = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
Answer:
(A) is 0.0773 mol B2H6
(C) is 2.79 x 10^23 H atoms
Explanation:
Questions (A) and (B) are the same.
2.14 g B2H6 x (1 mol B2H6/27.668g B2H6) = 0.0773 mol B2H6 (A)
<u>27.668 is the molar mass of B2H6 calculated from the period table: </u>
(2 x 10.81) + (6 x 1.008) = 27.668
1.008 is the mass of H and 10.81 is the mass of B
(C)
0.0773 mol B2H6 x (6 mol H/ 1 mol B2H6) x (6.022 x 10^23 H atoms/1 mol H)
= 2.79 x 10^23 hydrogen atoms
Further Explanation:
- For every 1 mol of B2H6, there are 6 moles of H (indicated by the subscript)
- 6.022 x 10^23 is Avogrado's number and it equals to 1 mol of anything
- Avogrado's number can be in units of atoms, molecules, or particles
The answer is A
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