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3 years ago

I NEED HELP PLEASE, THANKS! :)

Chemistry

1 answer:

Elan Coil [88]3 years ago

6 0

Answer:

Hi there!

Based on my own words, the electron cloud is a shell surrounding the nucleus that only consists of electrons.

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During science lab, some students added small pieces of magnesium ribbon to 10 mL of hydrochloric acid. They noticed that bubble

stiv31 [10]

Your answer would be a change in odor! Hope this helps! ;D

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3 years ago

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RSB [31]

Answer: Christine Herman & L.G Wade Jr., "2010". Organic Chemistry: Reaction of Alkane, 7e, Pearson Education, Radford University, Radford, VA.

Explanation:

This is an edited book. The Harvard reference style was used in the following order:

Authors name

Year of publication

Title

Edition

Publisher

Place of publication.

Note that the title of book should be italicized with capitalization of first word.

7 0

3 years ago

On the basis of the diagram, which of the following statements is true?

Furkat [3]

The answer is d I'm pretty sure because the second one and the first one don't make sense but I'm not 100 percent positive about it not being d ..... sorry if it's wrong.

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3 years ago

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ICMP scanning involves in checking for the live systems, which can be done by sending the following ping scan request to a host.

abruzzese [7]

Answer:

ICMP Echo Request

Explanation:

ICMP Echo Request is a form of probe or message sent by a user to a destination system.

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4 years ago

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

where,

$[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M$

Putting values in above equation, we get:

$pH=-\log(0.0731)\\\\pH=1.136$

Hence, the pH of the solution is 1.136

8 0

3 years ago