The ideal gas equation is;
PV = nRT; therefore making P the subject we get;
P = nRT/V
The total number of moles is 0.125 + 0.125 = 0.250 moles
Temperature in kelvin = 273.15 + 18 = 291.15 K
PV = nRT
P = (0.250 × 0.0821 )× 291.15 K ÷ (7.50 L) = 0.796 atm
Thus, the pressure in the container will be 0.796 atm
Answer:
b) 0.47
Explanation:
MwC5H12 = 72.15g/mol
⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12
MwC6H14=86.18g/mol
⇒molC6H14=(20.0)*(mol/86.18)=0,232
MwC6H6=78.11g/mol
⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6
<h3>XC6H14=(0.232)/(0.1386+0.232+0,128)=0.465≅0.47</h3>
Answer:
Drug calculation
If we have 45g of clobetasol = 0.05%w/w
Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g
It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w
Answer:
20 kJ/mol
Explanation:
From ∆G°= -RTlnK
But
Ag2SO4(s)<----------->2Ag+(aq) + SO4^2-(aq)
Ksp= [2Ag+]^2 [SO4^2-]
But Ag+ = 0.032M
Ksp= (2×0.032)^2 (0.032)
Ksp= 1.31072×10^-4
∆G°= -RTlnK
∆G°= -(8.314× 298×(-8.93976))= 20KJmol-1( to the nearest KJ)
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:
It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:
Thus, the mass in excess is:
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