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vazorg [7]
3 years ago
11

Which of the below will effect colligative properties the most? Be3N2 OK₂S CO Mgs

Chemistry
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

CO

Explanation:

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in dimesional analysis, if an electric vehicle gets 4 miles/ kWh and the utility company charges 12 cents per kWh for charging i
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Answer:

The cost of driving this car is 33.33 miles per dollar.

Explanation:

Distance covered in 1 kilo watt-hour = 4 miles/kWh

Utility company charges  = 12 cents/kWh

The cost of driving this car in miles per cents:

\frac{ 4 miles/kWh}{12 cents/kWh}=\frac{1 miles}{3 cents}

1 cents = 0.01$

The cost of driving this car in miles per dollars:

\frac{1}{3} miles/cent=\frac{1 miles}{3\times 0.01\$}=33.33 miles/\$

3 0
3 years ago
Which atom in the ground state has a stable valence electron configuration?
Alecsey [184]
(1) Ar is your answer. Argon is a noble gas, which is stable in its standard valence electron configuration because it already completed its valence shell following the octet rule.
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3 years ago
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The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper
LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{1/2}=\frac{1}{k\times a_0}

k = rate constant =?

a_0 = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a) Initial concentration of XY = a_o=0.100 M

Rate constant of the reaction = k = 6.96\times 10^{-3} M^{-1} s^{-1}

Half life of the reaction is:

t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}

=1,436.78 s

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = 6.20\times 10^{-2} M

\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become 6.20\times 10^{-2} M.

e)  Initial concentration of SO_2Cl_2= 0.050 M

Final concentration after time t = x

t = 55.0 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

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3 years ago
Which of the following changes has a decrease in entropy?
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In order to solve this problem you need to consider the states of matter and the amount of products.

If there are more products than reactants than there is an increase in entropy.

States of matter: increasing the motion of particles increases entropy.

Molecular movement:

Solid<Liquid<Gas

Changing states of matter from left to right would be an increase in entropy.

Changing states of matter from right to left would be a decrease in entropy

The answer is number 3

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3 years ago
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Which of the following is true about oxidation-reduction reactions?
zloy xaker [14]

Answer:

the last one probably

Explanation:

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