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IgorC [24]
3 years ago
12

What are the coefficients needed to balance the equation for the combustion of

Chemistry
2 answers:
Monica [59]3 years ago
6 0
CH
4
+ O
2
→
CO
2
+ H
2
O
First look at the C atoms. At first glance they are balanced with 1 C on each side.
Now look at the H atoms. They are not balanced. There are 4 H atoms on the left side and 2 H atoms on the right. Place a coefficient of 2 in front of the
H
2
O
. We now have 4 H atoms on both sides.
CH
4
+ O
2
→
CO
2
+ 2H
2
O
Now look at the O atoms. They are not balanced. There are 2 O atoms on the left side and 4 on the right. Place a coefficient of 2 in front of the
O
2
. We now have 4 O atoms on both sides.
CH
4
+ 2O
2
→
CO
2
+ 2H
2
O
The equation is now balanced. Each side has 1 C atom, 4 H atoms, and 4 O atoms.
Temka [501]3 years ago
6 0
Bnhhhhhhh ridurrei Hancock chick usato
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What happens to temperature and kinetic energy of particles during a change of state
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3 years ago
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

6 0
3 years ago
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