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kkurt [141]
3 years ago
9

Draw the lewis structure for the polyatomic trisulfide s−23 anion. be sure to include all resonance structures that satisfy the

octet rule.

Chemistry
1 answer:
mezya [45]3 years ago
3 0

The following are the steps involved in drawing Lewis structure of the polyatomic trisulfide anion S_{3}^{2-}:

Total number of valence electrons = (3 * 6) + 2 = 20 electrons

Resonance is not possible in this ion. The molecular geometry of the ion will be bent as there are two lone pairs and two bond pairs on the central atom.



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Why do you think the atomic number is basically represented by the number of protons and not the number of electrons?
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How does the number of subatomic particles change as the carbon isotope<br> decays?
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2 years ago
PLZ HELP!!
Nataly [62]

A and B are experiencing winter. The picture which isn't available here in this question is attached below.

Option C.

<h3><u>Explanation:</u></h3>

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Here in this diagram, we can see that the points A and B are the north pole and the part in northern hemisphere respectively which aren't facing the sun directly, whereas C and D are facing the sun. Thus the southern hemisphere is experiencing summer and the northern hemisphere the winter.

4 0
2 years ago
What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
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