Answer:
Partial pressure of CO₂ is 406.9 mmHg
Explanation:
To solve the question we should apply the concept of the mole fraction.
Mole fraction = Moles of gas / Total moles
We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)
Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles
To determiine the partial pressure of CO₂ we apply
Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P
Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure
We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
</span>
Answer:
30.33L
Explanation:
Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant and Charles law states that the volume of a given mass of gas is directly proportional to the temperature provided the pressure remains constant
P1V1/T1 = P2V2/ T2
P1 = 78atm, V1 = 21L , T1 = 900K
P2 = 45atm, V2 = ? , T2 =750K
78× 21 / 900 = 45×V2 / 750
1638/900 = 45 V2 / 750
1638×750 = 900×45V2
1228500 = 40500V2
Divide both sides by 40500
1228500÷40500= V2
V2 = 30.33L
I hope this was helpful, please mark as brainliest
I don’t know about 1 But number 2 is AB + xH2O = AB.xH2O
Typical day of a new day in my house with my phone in my car so I’m not going hard I don’t know how much it was
