Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
Answer:
The original concentration is "35 mg/ml".
Explanation:
According to the question,
The solution is diluted,
= 1:50
The initial volume,
V1 = 1 ml
Final concentration,
= 0.07 mg
then,
The final volume,
V2 = 500 ml
As we know,
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
OOH! I know photosynthesis and I have been doing it in school. So the Glucose and oxygen combine to help make the photosynthesis AND it is so important to not forget about Carbon Dioxide and all the features for that because a plant needs carbon dioxide while we need oxygen and plants give out oxygen so that humans can have it and we breathe out carbon dioxide so that plants can have it. We pretty much just help each other out. I plants weren't there than humans wouldn't be there. Hope this helps!
Molar mass H₂O = 18.0 g/mol
1 mol ----- 18.0 g
3.5 moles --- ?
Mass ( H₂O) = 3.5 x 18.0 / 1
=> 63.0 g
hope this helps!
