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Andre45 [30]
2 years ago
10

Which of the following statements, if true, would support the claim that the NO3− ion, represented above, has three resonance st

ructures? The NO3− ion is not a polar species. A The oxygen-to-nitrogen-to-oxygen bond angles are 90°. B One of the bonds in NO3− is longer than the other two. C One of the bonds in NO3− is shorter than the other two.
Chemistry
1 answer:
tester [92]2 years ago
7 0

Answer:

One of the bonds in nitrate is shorter than the other two.

Explanation:

We would firstly need to draw the Lewis structure for nitrate anion. To do this, let's follow the standard steps:

  • calculate the total number of valence electrons: five from nitrogen, each oxygen contributes 6, so a total of 18 from oxygen atoms, as well as one from the negative charge, we have a total of 24 valence electrons;
  • assign the central atom, usually this is the atom which is single; in this case, we have nitrogen as our central atom;
  • assign single bonds to all the terminal atoms (oxygen atoms);
  • assign octets to the terminal atoms and calculate the number of electrons assigned;
  • the number of electrons assigned is 24, so no lone pairs are present on nitrogen;
  • calculate the formal charges: each oxygen has a formal charge of -1 (formal charge is calculated subtracting the sum of lone pair electrons and bonds from the number of valence electrons of that atom); nitrogen has a formal charge of +2;
  • nitrogen doesn't have an octet as well, so we'll both minimize its formal charge and make it obtain an octet if we make one double bond N=O.

Therefore, we may have 3 resonance structures, as this double bond might be formed with any of the 3 oxygen atoms.

By definition, double bonds are shorter than single ones, so one of the bonds is shorter than the other two.

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How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin
faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water

6 0
2 years ago
A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200
Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

7 0
2 years ago
The density of gas particles in a section of Earth’s atmosphere decreases. Which of the following is the most likely explanation
Bad White [126]
<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
6 0
2 years ago
This information is taken directly from Exercise 3 in the CHEM 111/112 Laboratory Manual. Another example of excellence in the p
kolbaska11 [484]

Answer:

<h2>No</h2>

the information was not cited correctly....

Explanation:

I hope the following explanation will help you a lot.

5 0
2 years ago
Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
PilotLPTM [1.2K]

Answer:

k = 23045 N/m

Explanation:

To find the spring constant, you take into account the maximum elastic potential energy that the spring can support. The kinetic energy of the car must be, at least, equal to elastic potential energy of the spring when it is compressed to its limit. Then, you have:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the car = 1050 kg

k: spring constant = ?

v: velocity of the car = 8 km/h

x: maximum compression of the spring = 1.5 cm = 0.015m

You solve the equation (1) for k. But first you convert the velocity v to m/s:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant is 23045 N/m

3 0
1 year ago
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