The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.
<h3>What is Ideal law ?</h3>
According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.
PV = nRT.
Where,
- p = pressure
- V = volume (1.75 L = 1.75 x 10⁻³ m³)
- T = absolute temperature
- n = number of moles
- R = gas constant, 8.314 J*(mol-K)
Therefore, the number of moles is
n = PV / RT
State 1 :
- T₁ = (25⁰ C = 25+273 = 298 K)
- p₁ = 225 kPa = 225 x 10³ N/m²
State 2 :
- T₂ = 10 C = 283 K
- p₂ = 185 kPa = 185 x 10³ N/m²
The loss in moles of gas from state 1 to state 2 is
Δn = V/R (P₁/T₁ - P₂/T₂ )
V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N
p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)
p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)
Therefore,
Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))
= 0.0213 mol
Hence, The number of moles of gas lost is 0.0213 mol.
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Answer:
sodium hexachloroplatinate(IV)- Na2[PtCl6]
dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br
pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2
Explanation:
The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.
The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.
Answer:
<h2>1200 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 60 × 20
We have the final answer as
<h3>1200 J</h3>
Hope this helps you
X ml - <span>25% alcohol mixture
y ml - </span><span>90% alcohol mixture
x+y = 455
0.25x ml alcohol in </span>x ml of 25% alcohol mixture
0.9y ml alcohol in y ml of 90% alcohol mixture
0.75*455= 341.25 ml alcohol in 455 ml of 75% alcohol mixture
0.25x+0.9y=341.25
System of equations:
x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455
0.25x+0.9y=341.25
-0.25x-0.25y=-113.75
0.25x+0.9y=341.25 Add both equations
0.25x+0.9y-0.25x-0.25y=341.25-113.75
0.65y =227.5
y=227.5/0.65 = 350 ml of 90% alcohol mixture
x+y=455
x+355=455
x=100 ml of 25% alcohol mixture
