Answer:
Mass of CO2 WILL BE ~ 9.33 g
Explanation:
Moles of O2 = 14/18
Let the mass of CO2 be x
Then moles of CO2 will be = x/12
moles of CO2 = moles of O2
x/12 = 14/16
x = 9.33 grams
Ideal gas law PV = nRT or PM = dRt ( M is molecular mass of gas ) and d is gas density
where m/ M ( m is mass of gas sample )
m = dRT / P ( 2.78 g/L ) ( 0.082 L atm/ K mole )( 310K ) / 1.2 atm =
58.9 g/mol
gas must be butane ( C4H10) has molecular mass = 58.9 g/mol
<u>Answer:</u> The pressure of NO and 
in the mixture is 0.58 atm and 0.024 atm respectively.
<u>Explanation:</u>
We are given:
Equilibrium partial pressure of 
= 0.29 atm
For the given chemical equation:
Initial: a
At eqllm: a-2x 2x x
Calculating for the value of 'x'
Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm
Equilibrium partial pressure of = a - 2x = a - 2(0.29) = a - 0.58
The expression of 
for above equation follows:
We are given:
Putting values in above expression, we get:
Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.
So, 
Equilibrium partial pressure of = (a - 0.58) = (0.604 - 0.58) = 0.024 atm
Hence, the pressure of NO and in the mixture is 0.58 atm and 0.024 atm respectively.
Hello!
Saturated hydrocarbons are generally unreactive except in the presence of high temperatures and oxygen.
Saturated hydrocarbons are called Alkanes. These compounds are generally unreactive, but when exposed to high temperatures and oxygen they react in a combustion reaction to produce Carbon Dioxide and Water. This reaction is very violent and spontaneous and explosions of household gas cylinders occur because of the spontaneity of this reaction.
Have a nice day!
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
