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sveta [45]
2 years ago
14

A compound decomposes by a first-order process. If 37 % of the compound decomposes in 60 minutes, the half-life of the compound

is ________minutes.
Chemistry
1 answer:
professor190 [17]2 years ago
6 0

Answer:

90 minutes

Explanation:

The half life of a first order reaction is;

t1/2 = ln(2) / k

To solve for t;

The integral rate expression for a first order reaction is;

ln[A] = ln[Ao] - kt

t = 60 minutes

Final Concentration = [A] = 100 - 37 = 63

Initial Concentration = [Ao] = 100

ln(63) = ln(100) - k (60)

k (60) = ln (100) - ln(63)

k = 0.4620 / 60 = 0.0077

Inserting the value of k in the half life equatin;

t1/2 = ln(2) / 0.0077 = 90 minutes

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By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.

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3 years ago
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A student thinks that halogens are highly reactive because their electrons are weakly attracted to their nuclei. Which is eviden
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Answer: Halogens tend to attract electrons when bonding (Option C)

Explanation: Halogens being non metals have greater electronegativities hence, attract electrons and making the statement disputed. Nobel gases are highly stable; this explains why they are nonreactive. They do not form chemical bonds because they only have a little tendency to either gain or lose an electron; on the other hand, halogens are reactive because they only need one additional electron to complete their octet.  

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3 years ago
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Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. None of the above pairs will
Alla [95]

Answer:

Only Fe^{3+}+Mg gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential (E^{0}) of the reaction is positive. Because it leads to negative standard gibbs free energy change (\Delta G^{0}), which is a thermodynamic condition for spontaneity of a reaction.

E^{0}=E^{0}(reduction)-E^{0}(oxidation)

Where E^{0}(reduction) and E^{0}(oxidation) represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation:         Mg-2e^{-}\rightarrow Mg^{2+} ;  E_{Mg^{2+}\mid Mg}^{0}=-2.38V

Reduction:         Fe^{3+}+3e^{-}\rightarrow Fe ; E_{Fe^{3+}\mid Fe}^{0}=-0.04V

So, E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V

Hence this pair will give spontaneous reaction.

(2) Similarly as above, E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V

Hence this pair will give non-spontaneous reaction.

(4)  Similarly as above, E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V

Hence this pair will give non-spontaneous reaction.

6 0
3 years ago
Question 3 (5 points)
Daniel [21]

Answer:

332.918g O2

Explanation:

I'm having some issues with the work however, your final answer should be 332.918g O2

Hope this helped!

8 0
2 years ago
A 400 ml sample of gas is heated from -20 c to 60
Viktor [21]
Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
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7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml
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3 years ago
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