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olya-2409 [2.1K]
3 years ago
7

When h forms a bond with h2o to form hydronium ion, h3o , this bond is called a coordinate covalent bond because question 16 opt

ions: the oxygen no longer has eight electrons surrounding it. both bonding electrons come from the oxygen atom. it is an especially strong bond. the electrons are equally shared?
Chemistry
2 answers:
pshichka [43]3 years ago
7 0
The correct option would be the fact that both bonding electrons had come from the oxygen to form this chemical bond.

The other options are not true.
Anika [276]3 years ago
3 0

Both bonding electrons can’t me come from the oxygen atom.

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HELP WITH CHEMISTRY PLEASE!
maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

  • In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
  • Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

<u><em>In our problem:</em></u>

P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.


2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

<u><em>In our problem:</em></u>

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.


3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

<u><em>In our problem:</em></u>

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.


4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.


5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

<u><em>In our problem:</em></u>

P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

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Hydrogen Peroxide slowly decomposing into water and oxygen increases the entropy of the universe. Any reaction of a system always increases the degree of the orderliness of the universe. Decomposition is one of the best examples of increasing the entropy. This is when substance is broken down further to elements or other compounds.  
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