Answer:
The final temperature of hydrogen gas is 537.63 K.
Explanation:
Given data:
Initial volume = 2.00 L
Initial pressure = 740 mmHg (740/760 = 0.97 atm)
Initial temperature = 25 °C (25 +273 = 298 K)
Final temperature =?
Final volume = 3.50 L
Final pressure = standard = 1 atm
Formula:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂T₁ / P₁V₁
T₂ = 1 atm × 3.5 L × 298 K / 0.97 atm × 2.00 L
T₂ = 1043 atm .L. K / 1.94 atm. L
T₂ = 537.63 K
Explanation:
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Answer:
The temperature difference of the body after 3 hours = 5.16 K
Explanation:
we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹
or, 1.2 mole.h⁻¹
The average heat evolved by the oxidation of foodstuffs is then:
⇒ Q avg =
= 7.2 kj.h⁻¹.Kg⁻¹
the heat produced after 3 h would be:
= 7.2 kj. h⁻¹.Kg⁻¹ x 3 h
= 21.6 kj. kg⁻¹
= 21.6 x 10³ j kg⁻¹
We know Qp = Cp x ΔT
Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹
⇒ ΔT = 
⇒ ΔT = 
⇒ ΔT = 5.16 K
The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Given data:
Moles of hydrochloric acid = 0.385 mol
Mass of chlorine gas =?
Chemical equation:
4HCl + O₂ → 2Cl₂ + 2H₂O
Now we will compare the moles of Cl₂ with HCl.
HCl : Cl₂
4 : 2
0.385 : 2÷4× 0.385 = 0.1925 mol
Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.
Mass of Cl₂ :
Mass of Cl₂ = moles × molar mass
Mass of Cl₂ =0.1925 mol × 71 g/mol
Mass of Cl₂ = 13.6675 g
Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
Learn more about moles here:
brainly.com/question/8455949
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