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3 years ago

Each answer below describes the connectivity around a single atom.

Chemistry

1 answer:

MrRissso [65]3 years ago

4 0

Answer:

d. One single bond and two double bonds.

Explanation:

The octate rule is a chemical rule in which the atoms prefer to have eight electrons in the valence shell. Where a single bond provide two electrons and a double bond provide 4 electrons. Thus:

a. Two double bonds . Two double bonds provide 8 electrons. Octate rule is not violated

b. Three single bonds and one pair of electrons . Three single bonds provide 6 electrons and one pair of electrons provide two electrons. Thus, you have eight electrons and octate rule is not violated

c. Two single bonds and one double bond . Two single bonds provide four electrons and one double bond 4. Thus, you have eight electrons and octate rule is not violated.

d. One single bond and two double bonds. One single bond provides two electrons and two double bonds 8. Thus, you have 10 electrons and octate rule is violated.

e. Four single bonds. Four single bonds provide 8 electrons. Octate rule is not violated.

I hope it helps!

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Consider the following reaction. The system is allowed to reach an equilibrium and then FeS2(s) is added. What will happen if Fe

AveGali [126]

Answer:

More products (Fe2O3 and SO2) will be produced

Explanation:

In the reaction:

4 FeS2(s) + 11 O2(g) ⇌ 2 Fe2O3(s) + 8 SO2(g)

FeS2 is a reactant. When a reactant is added to a reaction which has already reached the equilibrium, more product is produced, so that the equilibrium is reestablished again

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3 years ago

The rate constant for a certain reaction is k = 4.50×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the

SIZIF [17.4K]

Answer: The concentration of reactant after the given time is 0.0205 M

Explanation:

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $4.50\times 10^{-3}s^{-1}$

t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)

$[A_o]$ = initial amount of the reactant = 0.400 M

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}$

$[A]=0.0205M$

Hence, the concentration of reactant after the given time is 0.0205 M

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3 years ago

What is the ph of a buffer consisting of 0.200 m hc2h3o2 and 0.200 m kc2h3o2? the k a for hc2h3o2 is 1.8×10−5. view available hi

AURORKA [14]

PH = pKa + log $\frac{[base]}{[Acid]}$
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case

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3 years ago

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Novosadov [1.4K]

Answer:

false I think, but not so sure

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3 years ago

Describe a light year. Help ASAP please, doing an exam..

Romashka [77]

Answer:

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Explanation:

Hope it helps

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