This problem is easily solvable because radioactivity equations are common and well-established. The pseudo-first reaction is written below:
A = A₀(1/2)^(t/h)
where
A is the final amount
A₀ is the original amount
t is the time
h is the half life
5,000 = A₀(1/2)^(24,000/6,000)
Solving for A₀,
<em>A₀ = 80,000 atoms</em>
The balanced equation for the reaction is as follows
2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂
stoichiometry of Al to H₂SO₄ is 2:3
number of Al moles reacted - 15.0 mol
if 2 mol of Al react with 3 mol of H₂SO₄
then 15.0 mol of Al reacts with - 3/2 x 15.0 mol = 22.5 mol
22.5 mol of H₂SO₄ is required
The answer is 3 It is a body of knowledge gained using inquiry and experimentation. Hope this helped!
Answer:
0.4
Explanation:
Given data:
Number of moles of SrCl₂ consumed = ?
Mass of ZnCl₂ produced = 54 g
Solution:
Chemical equation:
ZnSO₄ + SrCl₂ → SrSO₄ + ZnCl₂
Number of moles of ZnCl₂:
Number of moles = mass/ molar mass
Number of moles = 54 g/136.3 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of ZnCl₂ with SrCl₂ from balance chemical equation.
ZnCl₂ : SrCl₂
1 : 1
0.4 : 0.4
Thus when 54 g of ZnCl₂ produced 0.4 moles of SrCl₂ react.
Answer:
8L
Explanation:
Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant
P1V1= P2V2
P1 = 2atm, V1 = 12L ,
P2 = 3atm , V2 =
12 × 2 = V2 × 3
Divide both sides by 3
V2 = 24 ÷ 3
V2 = 8L
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