According of Dalton's law of Partial pressure, the total pressure of a mixture of gases is the sum of the partial pressures of the individual vases in the mixture.
Hence;
The for hydrogen collected over water, we have a mixture of hydrogen gas and water vapour.
Total pressure = pressure of hydrogen gas + vapour pressure of water
Pressure of hydrogen gas = Total pressure - vapour pressure of water
Pressure of hydrogen gas = 636 mmHg - 28.3 mmHg
Pressure of hydrogen gas = 607.7 mmHg
Answer:
25.11 g.
Explanation:
- It is clear from the balanced equation:
<em>Ag₂O + 2HCl → 2AgCl + H₂O.</em>
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that 1.0 mole of Ag₂O reacts with 2.0 moles of HCl to produce 2.0 mole of AgCl and 1.0 moles of H₂O.
- 7.8 g of HCl reacts with excess Ag₂O. To calculate the no. of grams of Ag₂O that reacted, we should calculate the no. of moles of HCl:
<em>no. of moles of HCl = mass/atomic mass</em> = (7.9 g)/(36.46 g/mol) = <em>0.2167 mol.</em>
- From the balanced equation; every 1.0 mol of Ag₂O reacts with 2 moles of HCl.
∴ 0.2167 mol of HCl will react with (0.2617 mol / 2 = 0.1083 mol) of Ag₂O.
<em>∴ The mass of reacted Ag₂O = no. of moles x molar mas</em>s = (0.1083 mol)(231.735 g/mol) = <em>25.11 g.</em>
Answer:
The balloon will occupy a volume of 6.48 L
Explanation:
<u>Step 1: </u>Data given
internal pressure = 1.00 atm
volume = 4.50 L
Temperature = 20.0 °C
<u>Step 2:</u> Calculate new volume via the ideal gas law
P*V = n*R*T
(P1*V1)/ T1 = (P2*V2)/T2
⇒ with P1 = 1.00 atm
⇒ with V1 = 4.50 L
⇒ with T1 = 20.0 °C = 293 Kelvin
⇒ with P2 = 0.600 atm
⇒ with V2 = TO BE DETERMINED
⇒ with T2 = -20°C = 253 Kelvin
(1.00atm * 4.50 L)/293 Kelvin = (0.600 atm*V2) / 253 Kelvin
0.01536 = 0.00237 V2
V2 = 6.48 L
The balloon will occupy a volume of 6.48 L
Answer:
this is too easy just do it yourself
