Search found 257 matches
- Fri Apr 08, 2022 9:21 am
- Forum: Chemistry forum
- Topic: Chemistry packages in 'R'
- Replies: 0
- Views: 35746
Chemistry packages in 'R'
What are the benefits of using eChem, chemCal and stoichcalc packages in 'R' ? I downloaded the R Documentations of all of the above mentioned packages. I executed some of the examples of R Documentations, user guides of all of the above mentioned packages. But I didn't understand the answers given ...
- Thu Mar 31, 2022 7:41 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13906
Re: Equilibrium
Correct answer, thanks.
- Wed Mar 30, 2022 6:09 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13803
Re: Equilibrium
That's a typographical mistake. I corrected it in my post.
- Wed Mar 30, 2022 5:05 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13906
Re: Equilibrium
I made appropriate changes in my posts as per your suggestions.
Now in answer to (c), There is an increase of 0.1 mol on SO3. As per chemical reaction 2SO2 + O2 →2 SO3, 50% of 0.10 mol increase in SO3 will be deducted from 0.20 mol of O2 and x is the proportion of an increase of 0.10 mol in SO3.
Now in answer to (c), There is an increase of 0.1 mol on SO3. As per chemical reaction 2SO2 + O2 →2 SO3, 50% of 0.10 mol increase in SO3 will be deducted from 0.20 mol of O2 and x is the proportion of an increase of 0.10 mol in SO3.
- Wed Mar 30, 2022 4:50 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13803
Re: Equilibrium
From ICE table, I found that we have to deduct 0.12 mol from 0.2 mol of F2 and 0.06 mol from 0.10 mol of Xe. That's why my answer matches with author's answer.
- Wed Mar 30, 2022 3:59 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13803
Re: Equilibrium
But author gave the answer 0.32 mol of F2 which is correct as per my calculations.
As per your calculation x=0.1 mol of F2
As per your calculation x=0.1 mol of F2
- Wed Mar 30, 2022 3:49 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13906
Re: Equilibrium
My answer to (c) is different than your answer.
My attempt to answer (c) =\(K=\frac{[SO_3]^2}{[SO_2]^2[O_2]}= \frac{ (1.6)^2}{(0.3 + 0.10x)^2(0.2-0.05)} \Rightarrow x=0.695041722823 mol SO_2/L\)
∴ (0.69 mol SO2/L)(10 L)= 6.9 mol of SO2
Is this answer correct? We have to add SO2.
My attempt to answer (c) =\(K=\frac{[SO_3]^2}{[SO_2]^2[O_2]}= \frac{ (1.6)^2}{(0.3 + 0.10x)^2(0.2-0.05)} \Rightarrow x=0.695041722823 mol SO_2/L\)
∴ (0.69 mol SO2/L)(10 L)= 6.9 mol of SO2
Is this answer correct? We have to add SO2.
- Wed Mar 30, 2022 12:22 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13906
Re: Equilibrium
(a)\(K= \frac{[SO_3]^2}{[SO_2]^2[O_2]}\)
(b)\(\frac{(\frac{15 mol}{10 L})}{(\frac{3 mol}{10 L})^2(\frac{2 mol}{10 L})}=125 \)
(c) I am working on this question.
(b)\(\frac{(\frac{15 mol}{10 L})}{(\frac{3 mol}{10 L})^2(\frac{2 mol}{10 L})}=125 \)
(c) I am working on this question.
- Tue Mar 29, 2022 11:56 pm
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13803
Re: Equilibrium
\((a)K= \frac{[XeF_4]}{[Xe][F_2]^2}= \frac{(0.10)}{(010)(0.20)^2}=25,\)
(b) 80% of 0.20 moles of Xe= 0.80 × 0.20 =0.16 mol XeF4
\(\frac{(0.16)}{(0.040)(0.20 + x -0.12)^2}=25\)
x=0.32 mol F2
(b) 80% of 0.20 moles of Xe= 0.80 × 0.20 =0.16 mol XeF4
\(\frac{(0.16)}{(0.040)(0.20 + x -0.12)^2}=25\)
x=0.32 mol F2
- Tue Mar 29, 2022 9:15 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13803
Equilibrium
Consider the gas-phase reaction Xe + 2F2 ⇌ XeF4 a) Calculate the equilibrium constant from the observation that at some temperature, the extent of reaction is 50% when 0.20 mol of Xe and 0.40 mol of F2 have been mixed in an empty 1.00-L-bulb. b)How many moles of F2 would have to be added to the equi...
- Tue Mar 29, 2022 8:23 am
- Forum: Chemistry forum
- Topic: Equilibrium
- Replies: 7
- Views: 13906
Equilibrium
The reaction of sulfur dioxide gas with oxygen gas to form sulfur trioxide gas is an important step in the production of sulfuric acid. a) Write the expression for the equilibrium constant. b)If a 10.0 L vessel contains 15 mol of SO3, 2.0 mol of O2, and 3.0 mol of SO2 at equilibrium at 1373 K, what ...
- Tue Mar 29, 2022 8:07 am
- Forum: Chemistry forum
- Topic: Stoichiometry
- Replies: 3
- Views: 10582
Re: Stoichiometry
a) Percentage composition of the compound:
Carbon:75.41%
Nitrogen:8.36%
Hydrogen:6.62%
Oxygen:9.61%
b) C21H22N2O2
c)C21H22N2O2
Carbon:75.41%
Nitrogen:8.36%
Hydrogen:6.62%
Oxygen:9.61%
b) C21H22N2O2
c)C21H22N2O2
- Tue Mar 29, 2022 2:27 am
- Forum: Chemistry forum
- Topic: Stoichiometry
- Replies: 1
- Views: 8991
Stoichiometry
One method for determining the percentage of nitrogen in a compound involves its quantitative conversion into nitrogen gas (N2), which is collected and measured. A 0.1000-g sample of a compound yielded 11.5 mL of N2. The density of N2 at the temperature and pressure at which the measurement was made...
- Tue Mar 29, 2022 2:01 am
- Forum: Chemistry forum
- Topic: Stoichiometry
- Replies: 3
- Views: 10582
Stoichiometry
The alkaloid strychnine, a compound containing carbon, hydrogen, nitrogen and oxygen was submitted for elemental analysis. Combustion of a 0.3432-g sample in oxygen gave 0.2032-g H2O and 0.9483-g CO2. A 0.4932-g sample yielded 36.5 ml of nitrogen gas with a density of 1.13 g/L. The molecular weight ...
- Mon Mar 28, 2022 6:27 am
- Forum: Chemistry forum
- Topic: Stoichiometry
- Replies: 8
- Views: 12718
Re: Stoichiometry
So, my answers to b), c) , d) are correct.
But you said equation in the answer a) is not balanced. How is that? Would you explain?
But you said equation in the answer a) is not balanced. How is that? Would you explain?