Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
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Answer:
i and ii
Explanation:
In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD
Balance Chemical Equation is as follow,
<span> Cu + 2 AgNO</span>₃ → 2 Ag + Cu(NO₃)₂
According to Balance Equation,
2 Moles of Ag is produced by reacting = 1 Mole of Cu
So,
0.854 Moles of Ag will be produced by reacting = X Moles of Cu
Solving for X,
X = (0.854 mol × 1 mol) ÷ 2 mol
X = 0.427 Moles of Cu
Result:
0.854 Moles of Ag are produced by reacting 0.427 Moles of Cu.
The change of phase from a vapor to a liquid is called condensation.
Hope this helps~
Answer:
Zn 3
s 1
o 4
Zn. zine
s. sulphate
o oxygen
3znso4. zine sulphateoxide
