HI
So, the formula for water is H2O
When you have the same amount of the reactants , hydrogen will be the limiting reactant.
Limiting reactant is the thing that runs out first.
Answer:
qualitative data : quantitative data :
circular in shape 75 colonies ...
stained purple 200 purple ..
spreads across plate 55 colonies ...
Explanation:
i got it right :) .
Answer:
H₂SO₄
Explanation:
Given data:
Number of moles of H₂SO₄ = 15 mol
Number of moles of Fe = 13 mol
Which reactant is limiting reactant = ?
Solution:
Chemical equation:
3H₂SO₄ + 2Fe → Fe₂(SO₄)₃ + 3H₂
now we will compare the moles reactant with product.
H₂SO₄ : Fe₂(SO₄)₃
3 : 1
15 : 1/3×15 = 5
H₂SO₄ : H₂
3 : 3
15 : 15
Fe : Fe₂(SO₄)₃
2 : 1
13 : 1/2×13 = 6.5
Fe : H₂
2 : 3
13 : 3/2×13 = 19.5
Number of moles of product formed by H₂SO₄ are less thus it will act as limiting reactant.
The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months
To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Amount remaining (N) = 5%
Original amount (N₀) = 100%
<h3>Number of half-lives (n) =?</h3>
N₀ × 2ⁿ = N
5 × 2ⁿ = 100
2ⁿ = 100/5
2ⁿ = 20
Take the log of both side
Log 2ⁿ = log 20
nlog 2 = log 20
Divide both side by log 2
n = log 20 / log 2
<h3>n = 4.32</h3>
Thus, 4.32 half-lives gas elapsed.
Finally, we shall determine the half-life of the element. This can be obtained as follow.
Number of half-lives (n) = 4.32
Time (t) = 500 years
<h3>Half-life (t½) =? </h3>
t½ = t / n
t½ = 500 / 4.32
t½ = 115.74 years
Multiply by 12 to express in months
t½ = 115.74 × 12
<h3>t½ ≈ 1389 months </h3>
Therefore, the half-life of the radioactive element in months is approximately 1389 months
Learn more: brainly.com/question/24868345
