It's very simple... if we remember value of Universal Gas Constant R and Ideal Gas Law, so...
Ideal Gas Law
pV = nRT, where:
p - pressure (in kPa),
V - volume (in L),
n - number of moles (in mol),
R - universal cas constant (in kPa * L / mo l* K),
T - temperature (in K)
n = m/M, where:
n - number of moles,
m - mass (in grams),
M - molar mass of ingredient (in g/mol) - you find this at Periodic Table.
pV = nRT ---> pV = mRT/M ---> pVM = mRT ---> pVM/RT = m
p = 17615 kPa
T = 273.15 + 23 = 296.15 K
V = 43.8 L
R = 8.314 kPa * L / mol * K
M (for argon) = 39.948 g/mol
and
m = (17615 kPa * 48.3 L * 39.948 g/mol) / (296.15 K * 8.314 kPa * L / mol * K)
m = 13803.93 grams of Argon
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;
= 0.8
The rate-out
=
=
We can say that:
where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12
Integration of the above linear equation =
so we have:
∴
Since A(0) = 12
Then;
Hence;
∴ the concentration at 10 minutes is ;
= %
= 0.0456667 %
= 0.046% to three decimal places
Answer:
Explanation:
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Answer:
18.58 moles of Cu
Solution:
Data Given:
Chemical Formula of Tennantite = Cu₁₂As₄S₁₃
Mass of Cu₁₂As₄S₁₃ = 2290 g
M.Mass of Cu₁₂As₄S₁₃ = 1479.06 g.mol⁻¹
Step 1: Calculate Moles of Cu₁₂As₄S₁₃ as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 2290 g ÷ 1479.06 g.mol⁻¹
Moles = 1.548 mol
Step 2: Calculate Moles of Cu,
As,
1 mole of Cu₁₂As₄S₁₃ contains = 12 moles of Cu
So,
1.548 mol of Cu₁₂As₄S₁₃ will contain = X moles of Cu
Solving for X,
X = (1.548 mol × 12 mol) ÷ 1 mol
X = 18.58 moles of Cu
Answer:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Explanation:
A → B + C
The rate law of the reaction will be :
Initial rate of the reaction when concentration of the substrate was 0.4 M:
..[1]
Initial rate of the reaction when concentration of the substrate was 0.8 M:
...[2]
[1] ÷ [2] :
x = 0
The order of the reaction is zero.
For the value of rate constant ,k:
..[1]
x = 0
k= 0.183 M/s
The half life of the zero order kinetics is given by :
Where:
= Initial concentration of A
k = Rate constant of the reaction
So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.