When a weight is reciprocated, the equal and opposite force required for its acceleration at any instant appears as an unbalanced<span> force on the frame of the machine to which the weight belongs.</span><span>
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Given 1 gm of MCl2 fors 1.286g of AgClmolar mass of AgCl is 143.5 g 1.286g of AgCl contains 1.286/143.5 mols of AgCl =0.0089616 molseach mol of MCl2 release two Cl- ionshence number of mols MCl2 given(1gm) is 0.0089616 /2=0.0044808 mols hnce one mol of MCl2 weighs 1/0.0044808= 223.17g
So it appears that this is asking for what the volume (V) of the N2 gas is under those conditions. For this we use the Ideal Gas Law:
PV = nRT, where P = the pressure in atm, V = the volume in liters (L), n = the number of moles of gas, R = the Ideal gas constant, which is 0.0821 (atm•L)/(mol•K), and T = the temperature in Kelvin which is the °C + 273
So to solve for V: V = (nRT)/P
V = [(0.020molN2)×(0.0821atm•L/(mol•K)×290K] ÷ 1.5atm
V = [(0.476•atm•L N2] ÷ 1.5atm
V = 0.317 L N2
If you look at each step, you can see that I included the units, because you can see them cross each other out to give you the final product, volume, which is in liters (L).
I hope that helps you understand these types if problems better!
Answer:
V₂ = 19.3 L
Explanation:
Given data:
Initial volume = 29.3 L
Initial pressure = standard = 1 atm
Initial temperature = standard = 273 K
Final temperature = -23°C (-23+273 = 250K)
Final volume = ?
Final pressure = 1.39 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1 atm × 29.3 L × 250 K / 273 K × 1.39 atm
V₂ = 7325 atm .L. K / 379.47 K.atm
V₂ = 19.3 L
The only one I know for sure is Mass is always conserved In a Chemical reaction.