I believe the balanced chemical equation is:
C6H12O6 (aq) + 6O2(g)
------> 6CO2(g) + 6H2O(l)
First calculate the
moles of CO2 produced:
moles CO2 = 25.5 g
C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)
moles CO2 = 0.8493 mol
Using PV = nRT from
the ideal gas law:
<span>V = nRT / P</span>
V = 0.8493 mol *
0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm
<span>V = 22.28 L</span>
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Number 1 is incorrect, the genotypes are given to you. You need to use GG and gg. The outcome would be 100% Gg.
Number 2 is incorrect, the genotypes are given to you. You need to use Gg and Gg. The outcome would be 25% gg, 25% GG, and 50% Gg.
Number 3 is incorrect, the genotypes are given to you. You need to use TT and tt. The outcome would be 100% Tt.
Number 4 is incorrect, the genotypes are given to you. You need to use RR and rr. The outcome would be 100% Rr.
Please read the directions and use the genotypes they give you! The information is all there for you, you just need to put it in correctly. If you're still having trouble understanding how to do this, feel free to message me and I'd be happy to help you.
Answer: 4.41 atm
Explanation:
Given that,
Original pressure of oxygen gas (P1) = 5.00 atm
Original temperature of oxygen gas (T1) = 25°C
[Convert 25°C to Kelvin by adding 273
25°C + 273 = 298K
New pressure of oxygen gas (P2) = ?
New temperature of oxygen gas (T2) = -10°C
[Convert -10°C to Kelvin by adding 273
-10°C + 273 = 263K
Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law
P1/T1 = P2/T2
5.00 atm /298K = P2/263K
To get the value of P2, cross multiply
5.00 atm x 263K = 298K x V2
1315 atm•K = 298K•V2
V2 = 1315 atm•K / 298K
V2 = 4.41 atm
Thus, the new pressure inside the canister is 4.41 atmosphere
