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blsea [12.9K]
3 years ago
6

A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m

oles of CO2 would be formed if 3moles of the compound react with oxygen?
Chemistry
1 answer:
Phoenix [80]3 years ago
7 0
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
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3 years ago
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When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.
Lubov Fominskaja [6]

Answer:

True

Explanation:

When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.

3 0
3 years ago
When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reacti
Delvig [45]

Answer:

CH4+2O2 = CO2+2H2O

7 0
3 years ago
Question 7
Paladinen [302]

Answer:

2HCI + Na2S - > H2S + 2NaCl

Explanation:

2HCI + Na2S -> H2S + 2NaCl

before reaction: 2H, 2Cl, 2Na, 1S

after reaction: 2H, 2Cl, 2Na, 1S

8 0
4 years ago
What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29
hram777 [196]

Henderson–Hasselbalch equation is given as,

                                         pH  =  pKa  +  log [A⁻] / [HA]   -------- (1)

Solution:

Convert Ka into pKa,

                                         pKa  =  -log Ka

                                         pKa  =  -log 1.37 × 10⁻⁴

                                         pKa  =  3.863

Putting value of pKa and pH in eq.1,

                                         4.29  =  3.863 + log [lactate] / [lactic acid]

Or,

                   log [lactate] / [lactic acid]  =  4.29 - 3.863

                   log [lactate] / [lactic acid]  =  0.427

Taking Anti log,

                             [lactate] / [lactic acid]  =  2.673

Result:

           2.673 M  lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.

6 0
4 years ago
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