Answer:
B) 16 g
Explanation:
First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4 mol O₂ *
= 8 mol H₂
Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:
- 8 mol H₂ * 2 g/mol = 16 g
Thus, the correct answer is option B).
464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O
mol of NO = </span>
=
= 0.93 mol
Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3
If mol of NO = 0.93 mol
then mol of NH₃ =
= 0.62 mol
Mass of ammonia = mol × molar mass
= 0.62 mol × 17.03 g/mol
= 10.54 g
Therefore B is the best answer
Answer:
k = -0.0525 s⁻¹
Explanation:
The equaiton for a first order reaction is stated below:
ln[A]=−kt+ln[A]₀.
[A] = 5.50 x 10⁻³ M
[A]₀ = 7.60 x 10⁻² M
t = 85.0 - 35.0 = 50.0 s
The rate constant is represented by k and can be calculated substituting the values given above:
k = (ln[A]₀ - ln[A])/t
k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s
k = -0.0525 s⁻¹
The sugar solution was made by mixing 12 grams of water
