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:
Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.
Number of half lives in 9612 years = 9612/1602 = 6 half lives
New mass = Original mass x (1/2)n where n is the number of half lives.
Therefore, New mass= 500 x (1/2)∧6
= 500 x 0.015625
= 7.8125 g
Hence the mass of radium after 9612 years will be 7.8125 grams.
Explanation:
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Answer:
Number of moles = 10.6 mol
Explanation:
Given data:
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)
Number of moles of citric acid = ?
Solution:
Formula:
Number of moles = mass/molar mass
Now we will calculate the molar mass of citric acid:
C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)
C₆H₈O₇ = 72.06 + 8.064+112
C₆H₈O₇ = 192.124g/mol
Number of moles = 2030 g/ 192.124g/mol
Number of moles = 10.6 mol
Answer:
a.
△H=−72 kcal
The energy required for production of 1.6 g of glucose is [molecular mass of glucose is 180 gm]
b.
The iron(III) ions and chloride ions remain aqueous and are spectator ions in a reaction that produces solid barium sulfate.
