A laboratory report is an important part of the scientific process that communicate the important work that has been done.
<h3>How to depict the laboratory report?</h3>
Your information is incomplete. Therefore, an overview of a laboratory report will be given. The laboratory report is important for future studies a d experiments.
A laboratory report is broken down intosections such as title, abstract, introduction, methods, materials, results, discussion conclusion, and references.
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Answer:
The initial volume in mL is 5959.2 mL
Explanation:
As the number of moles of a gas increases, the volume also increases. Hence, number of moles and volumes are directly proportional i.e
n ∝ V
Where n is the number of moles and V is the volume
Then, n = cV
c is the proportionality constant
∴n/V = c
Hence n₁/V₁ = n₂/V₂
Where n₁ is the initial number of moles
V₁ is the initial volume
n₂ is the final number of moles
and V₂ is the final volume.
From the question,
n₁ = 0.693 moles
V₁ = ?
n₂ = 0.928 moles
V₂ = 7.98 L
Putting the values into the equation
n₁/V₁ = n₂/V₂
0.693 / V₁ = 0.928 / 7.98
Cross multiply
∴ 0.928V₁ = 0.693 × 7.98
0.928V₁ = 5.53014
V₁ = 5.53014/0.928
V₁ = 5.9592 L
To convert to mL, multiply by 1000
∴ V₁ = 5.9592 × 1000 mL
V₁ = 5959.2 mL
Hence, the initial volume in mL is 5959.2 mL
Answer:
Yea.....where is the article and the 6 questions?
Answer:
19.4 g of alum, will be its theoretical yield
Explanation:
The reaction is:
2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O
Let's determine the amount of acid.
M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution
M = mmol /mL
M . mL = mmol
We replace: 8.3 mL . 9.9 M = 82.17 mmoles
We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles
Ratio is 4:2
4 moles of sulfuric acid can make 2 moles of alum
By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.
We convert moles to mass:
Molar mass of alum is: 473.52 g/mol.
0.041085 moles . 473.52 g/mol = 19.4 g
Answer: 1.59atm
Explanation:
We have that for the Question "Calculate the final pressure of the gas mixture, assuming that the container volume does not change."
it can be said that
The final pressure of the gas mixture, assuming that the container volume does not change =
From the question we are told
A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely decomposes, producing NO2(g) and NO(g).
