The time required to reduce the concentration from 0.00757 M to 0.00180 M is equal to 1.52 × 10⁻⁴ s. The half-life period of the reaction is 9.98× 10⁻⁵s.
<h3>What is the rate of reaction?</h3>
The rate of reaction is described as the speed at which reactants are converted into products. A catalyst increases the rate of the reaction without going under any change in the chemical reaction.
Given the initial concentration of the reactant, C₀= 0.00757 M
The concentration of reactant after time t is C₁= 0.00180 M
The rate constant of the reaction, k = 37.9 M⁻¹s⁻¹
For the first-order reaction: 
0.00180 = 0.00757 - (37.9) t
t = 1.52 × 10⁻⁴ s
The half-life period of the reaction: 
Half-life of the reaction = 9.98 × 10⁻⁵s
Learn more about the rate of reaction, here:
brainly.com/question/13571877
#SPJ1
Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Answer:
9.72 grams.
Explanation:
From the equation, 4 moles of NH₃ produce 6 moles of water.
Therefore the reaction to product ratio of NH₃ to H₂O is 4:6
and 2:3 into its simplest form.
The number of moles of NH₃ in 6.12 g is:
Number of moles=mass/ RMM
=6.12 g/17 G/mol
=0.36 moles.
Therefore the number of moles of H₂O produced is calculated as follows.
(0.36 Moles×3)2 = 0.54 moles
Mass= Number of moles × RMM
=0.54 moles×18g/mol
=9.72 grams.
Answer:
Creo que la respuesta es n. Transalation : I think the answer is N.
Explanation:
