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11111nata11111 [884]
3 years ago
14

A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O.

What is the empirical formula of this CHO compound? Enter as C#H#O#, e.g. C2H3O2

Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Answer: the empirical formula is CH2O

Explanation:Please see attachment for explanation

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If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).​
11111nata11111 [884]

Answer:

{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

7 0
3 years ago
Why is it important for organelles to double during the G2 phase of the cell cycle?
boyakko [2]

Explanation:

so then we would know about the reproduction system and how we are born and what happens with the cells inside the womens womb and how the sperm cell meets the egg cell and creates the egg

6 0
2 years ago
Which of the following is is a chemical property of pure water
german

Answer:

Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale

5 0
2 years ago
Read 2 more answers
126785033% Divided by 348675948849 = <br> what does it equal
meriva

Answer:

0.00000363618

could be wrong.

double check me someone or just trust me

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5 0
3 years ago
Free Energy
ratelena [41]

Answer:

a) galvanic cell

b)electrolytic cell

c) i) K=6.27x10'34

ΔG°=198790 J

ii) K=3.58x10'-34

ΔG°= 191070 J

d) E°=0.278 v

ΔG°= -26827 J

Explanation:

a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".

The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.

b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.

c) i) look image attached

ii) k = look image attached

ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)

ΔG° =-191070

d) E°= 0.0592 v/n x lg K

E°= 0.0592V / 1 X log 5.0X10'4

E°= 0.278 v

ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v

ΔG° = -26827 J

5 0
3 years ago
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