The reaction of removing CO2
using LiOH is the following:
2 LiOH + CO2 -----> Li2CO3
+ H2O
By solving the amount of CO2
the LiOH can scrub:
(3.50 × 10^4 g LiOH) (1 mol LiOH/
24 g LiOH) ( 1 mol CO2 / 2 mol LiOH) ( 44 g CO2 /1 mol CO2) = 32, 083.33 g CO2
it can scrub
<span>Since number of astronaut = 32,
083.33 g / 9 (8.8 × 10^2) = 4 astronaut</span>
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
Um i think gold... i think?
<span>All bulbs/plants and creatures </span>want nitrogen<span> to produce amino hallucinogen, proteid and DNA, although this </span>nitrogen<span> in the air is not in a manner that everybody can use.
For more information, look at the appendage below!</span>
C I think it’s C I’m semi guessing
