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4 years ago

How do the resonance structures for ozone, 03, differ?

Chemistry

1 answer:

Anni [7]4 years ago

3 0

Explanation:

Ozone, or 03,has two major resonance structures that contribute equally to the overall hybird structure of the molecule. The two structures are equivalent from the stability standpoint, each having a positive and a negative formal charge placed on two of the oxygen atoms.

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Hello there!

Alenkasestr [34]

Answer:

Mass of carbon dioxide = 7.48 g

Explanation:

Given data:

Mass of lithium carbonate = 12.5 g

Mass of carbon dioxide produced = ?

Solution:

Chemical equation:

Li₂CO₃ → Li₂O + CO₂

Number of moles of Li₂CO₃:

Number of moles = mass/ molar mass

Number of moles = 12.5 g /73.89 g/mol

Number of moles = 0.17 mol

Now we will compare the moles of Li₂CO₃ with CO₂.

Li₂CO₃ : CO₂

1 : 1

0.17 : 0.17

Mass of carbon dioxide:

Mass of carbon dioxide = number of moles × molar mass

Mass of carbon dioxide = 0.17 mol × 44 g/mol

Mass of carbon dioxide = 7.48 g

4 0

3 years ago

An experiment was undertaken to determine the effect of different liquids upon the swelling of a vulcanized sample of natural ru

N76 [4]

Answer:

a. 459.86 gmol

b. 0.591

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

7 0

3 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

3 years ago

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Svet_ta [14]

Answer:

Jsldjafjlas;jdiljf;laemjkijsd;fjakle

Explanation:

Kidman;lkjaweko no[an[sdomk’m kjioasjpoie;nojnaodjflkadm;oaiwjdaWMDPAIklsdwiaskdljwiojda;JSDJW;OIJ;OAJAWIODJ;jwoifpajsdok;thdyvajewoz,fox’s,dflxdsmxfew,Pom;f

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7 0

3 years ago

Find the ph of a buffer that consists of 0.18 m ch3nh2 and 0.73 m ch3nh3cl (pkb of ch3nh2 = 3.35)?

ozzi

Hello!

First, we need to determine the pKa of the base. It can be found applying the following equation:

$pKa=14-pKb=14-3,35=10,65$

Now, we can apply the Henderson-Hasselbach's equation in the following way:

$pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04$

So, the pH of this buffer solution is 10,04

Have a nice day!

8 0

3 years ago