From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.
<h3>What is combustion?</h3>
Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)
We can obtain the number of moles of CO2 from;
PV = nRT
n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K
n = 7.29 /32.6
n = 0.22 moles
If 6 moles of oxygen produces 4 moles of CO2
x moles of oxygen produces 0.22 moles of CO2
x = 0.33 moles
1 mole of oxygen occupies 22.4 L
0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole
= 7.4 L of oxygen
Learn more about stoichiometry: brainly.com/question/13110055
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T₁ = 40°C + 273.15 = 313.15 Kelvin T₂ = 30°C + 273.15 = 303.15 Kelvin
Solving Gay-Lussac's Law for P₁ we get:
P₁ = P₂ • T₁ ÷ T₂ P₁ = 760 torr • 313.15 K ÷ 303.15 K P₁ = 785.07 torr
Using the calculator, we click on the P1 button.
We then enter the 3 numbers 760 313.15 and 303.15 into the correct boxes then click "CALCULATE" and get our answer of 785.07 torr.
According to the reaction equation:
CH3COO- + H+ → CH3COOH
initial 0.25 0.15
change - 0.025 + 0.025
Equ (0.25-0.025) (0.15 + 0.025)
first, we have to get moles acetate and moles acetic acid:
moles of acetate = 0.25 - 0.025 = 0.225 moles
∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M
moles of acetic acid = 0.15 + 0.025 = 0.175 moles
∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M
Pka = -㏒ Ka
= -㏒ 1.8 x 10^-5
= 4.74
from H-H equation we can get the PH value:
PH = Pka + ㏒ [acetate / acetic acid]
PH = 4.74 + ㏒[0.225/0.175]
∴ PH = 4.8
Answer:
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