K1x1=k2x2=F
k2x2/k1=x1
F=keqx=keq(x1+x2)
sub.s in for x1:F=keq(k2x2/k1+x2)
sub.s in for F (in terms of k2 and x2)k2x2=keq(k2x2/k1+x2)
x2 is in every term:k2=keq(k2/k1+1)
multiply both sides by (k1/k2+k1)
keq=(k1/k1+k2)k2
keq=(1/k1+1/k2)−1
Answer:
The answer to your question is 6.77 atm
Explanation:
Data
Pressure 1 = P1 = 7.5 atm
Temperature 1 = T1 = 65°C
Pressure 2 = P2 = ?
Temperature 2 = T2 = 32°C
Process
-Use Gay-Lussac law to solve this problem
P1/T1 = P2/T2
-Solve for P2
P2 = P1T2 / T1
-Convert temperature to °K
T1 = 65 + 273 = 338°K
T2 = 32 + 273 = 305°K
-Substitution
P2 = (7.5 x 305) / 338
-Simplification
P2 = 2287.5 / 338
-Result
P2 = 6.77 atm
Answer:
0.98 g of H₂
Explanation:
the balanced equation for the reaction is
Mg + 2HCl ---> MgCl₂ + H₂
molar ratio of Mg to H₂ is 1:1
number of Mg moles reacted = 12 g/ 24.3 g/mol = 0.49 mol
according to molar ratio
when 1 mol of Mg reacts 1 mol of H₂ is formed
therefore when 0.49 mol of Mg reacts - 0.49 mol of H₂ forms
therefore mass of H₂ formed = 0.49 mol x 2 g/mol = 0.98 g
mass of H₂ formed is 0.98 g
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows:
Mg=24 g/mole
Cl=38 g/mole
Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so:
(38 g/mole*2 moles)+24 g/mole=100 g/mole
Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows:
3.00 moles * 100 g/mole = 300 g</span>
