Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
The masses of CO and CO2 are 90.55g and 100−90.55=9.45 g respectively.
<h3>Total mass.</h3>
Let the mixture has 100g as total mass.
The number of moles of CO is 2890.55=3.234.
The number of moles of CO2 is 449.45=0.215.
The mole fraction of CO is 3.234+0.2153.234=0.938.
The mole fraction of CO2 is 1−0.938=0.062.
The partial pressure of CO is the product of the mole fraction of CO and the total pressure.
It is 0.938×1=0.938 atm.
The partial pressure of carbon dioxide is 0.062×1=0.042 atm.
The expression for the equilibrium constant is:
Kp=PCO2PCO2=0.062(0.938)2=14.19
Δng=2−1=1
Kc=Kp(RT)−Δn=14.19×(0.0821×1127)−1=0.153.
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The mass of an element listed in the Periodic Table is the weighted average of all its naturally occurring isotopes.
Naturally occurring carbon is about
99 % carbon-12 (12.000 u) + 1 % carbon-13 (13.003 u).
That extra carbon-13 makes the <em>average atomic mass</em> greater than 12.000 u.
5.5 moles
2.0=x/2.75
2.0*2.75= x/2.75 (2.75)
5.5=x
