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Questions (6)

When CO2(g) (0.009242 mol/L) and 0.009242 mol/L of H2(g) in a 460.0 L reaction vessel at 619.0 °C are allowed to come to
1. 3 answers
2. 444 views
When CH4(g) (0.06318 mol/L) and 16.43 mol of H2S(g) in a 130.0 L reaction vessel at 711.0 °C are allowed to come to equilibrium
1. 5 answers
2. 565 views
When a sample of NH3(g) (296.5 grams) is placed in 140.0 L reaction vessel at 676.0 °C and allowed to come to equilibrium the
1. 3 answers
2. 487 views
When F2(g) (0.02028 mol/L) and 0.02028 mol/L of Cl2(g) in a 460.0 L reaction vessel at 774.0 K are allowed to come to
1. 3 answers
2. 794 views
When a sample of I2(g) (0.07249 mol/L) is placed in 130.0 L reaction vessel at 865.0 K and allowed to come to equilibrium the
1. 5 answers
2. 636 views
When a sample of I(g) (18.15 mol) is placed in 59.00 L reaction vessel at 605.0 °C and allowed to come to equilibrium the
1. 4 answers
2. 558 views

.04107 x 130 = 5.3391 4x= 5.3391 x = 1.3348 therefore H2S(g) = 1.3348/130 = .01027?
is this simply x = -1.327 therefore .009242-(-1.327) = 1.3362 then 1.3362 / 460.00 = .0029?
thank you dr. bob!
i have no clue on how to set this up....
is this simply 3x=6.250 x= 2.083 therefore 2.083/140 = .01489?
.............F2(g) + Cl2(g) ==> 2ClF(g) I........0.02028 ...... 0.02028 ........ 0 C...........x....................x.............. -2x E.....0.02028+x ..... 0.02028+x...... -2x 2x = 0.02898 = x = .01449 therefore 0.02028 - .01449 = .00579?
Then mols I2 = 0.07249-0.0555 = 0.0170 mols correct? therefore, m I2 = 0.0170/130= 1.308 x 10^-4 ??
The problem is I still don't understand the concept and what is being asked... Anything you can recommend DrBob?? I would appreciate it as I would like to catch up.... Ok so .............I2(g) ==> 2I I............ .07249 .......0
My issue is that I took the first semester of general chemistry almost 2 years ago. I have forgotten everything. This is why everything is brand new to me... Can you recommend any online reading to catch up? So for this problem this is what I get... Please

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