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Helen [10]
2 years ago
5

Which molecule will have a double covalent bond? #3 #5 & #6

Chemistry
1 answer:
KatRina [158]2 years ago
4 0

Answer:the CO2 molecule has an excess of electron

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The orbital period of planet Venus is 0.62 years. What is its distance from the sun?
Vikki [24]

Kepler's third law shows the relationship between the orbital period of an object and the distance between the object and the object it orbits.

The simplified version of this law is: P^2 = a^3

Where,

P = period of the orbit in years = 0.62 years

a = average distance from the object to the object it orbits in AU. The astronomical unit AU is a unit of length which is roughly equivalent to the distance from Earth to the Sun.

Therefore calculating for a:

0.62 ^ 2 = a ^ 3

a = 0.62 ^ (2/3)

a = 0.727 AU = 0.72 AU

Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to Sun.

<span>Answer: B. 0.72 AU</span>

5 0
3 years ago
Which of these is an example of a physical change? Question options: melting a substance breaking chemical bonds forming chemica
kherson [118]
Your answer is Breaking chemical bonds. Hope this helps
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2 years ago
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Saul converts 2.392 hectoliters to liters. What should his new number be?
Lostsunrise [7]
2.392 hectoliters = 239.2 liters. 1 hectoliter = 100 liters.
3 0
3 years ago
Give one example of fast erosion and one example of slow erosion. Use complete sentences.
topjm [15]
An example of erosion is the Grand Canyon, which was worn away over time by the Colorado river.
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3 years ago
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Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
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