The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.
<h3>What is Combined gas law?</h3>
Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.
It is expressed as;
P₁V₁/T₁ = P₂V₂/T₂
Given the data in the question;
- Initial volume V₁ = 14.5L
- Initial pressure P₁ = 0.980atm
- Initial temperature T₁ = 20.0°C = 293.15K
- Final pressure P₂ = 740.mmHg = 0.973684atm
We substitute our given values into the expression above.
P₁V₁/T₁ = P₂V₂/T₂
( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂
14.21Latm / 293.15K = 13.92368Latm / T₂
14.21Latm × T₂ = 13.92368Latm × 293.15K
14.21Latm × T₂ = 4081.72679LatmK
T₂ = 4081.72679LatmK / 14.21Latm
T₂ = 287.24K
T₂ = 14.09°C
Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.
Learn more about the combined gas law here: brainly.com/question/25944795
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Answer:
See BELOW!
Explanation:
Your question is confusing, you are not mentioning specifically what you are trying to balance the equation of. If you require more assistance, write in the comments and I'll be glad to assist. you!
Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
You must use 64.43 g H₂O.
<em>Balanced chemical equation</em>: H₂O + CO₂ → H₂CO₃
<em>Moles of CO₂</em> = 157.35 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 3.5753 mol CO₂
<em>Moles of H₂O</em> = 3.5753 mol CO₂ × (1 mol H₂O/1 mol CO₂) = 3.5753 mol Fe
<em>Mass of H₂O</em> = 3.5753 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 64.43 g H₂O
