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3 years ago

13

What is the maximum number of electrons in the following energy level?

1 answer:

Licemer1 [7]3 years ago

3 0

Energy level 1, shell letter K n electron capacity 2
Energy level 2, shell letter L and electron capacity 8
Energy level 3, shell letter M and election capacity 18
Energy level 4, shell letter N and electron capacity 32

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Explain what might happen to tundra animals, such as polar bears, as earth's climate warms. pls explain

SCORPION-xisa [38]

Answer:

They will die.

Explanation:

Polar Bears were made to live in the frigid temperatures of the tundras. If there is a temperature change it can affect their habitat. It can destroy it and they would have to move out to another place. They can die due to global warming.

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3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

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

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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3 years ago

Perform the following mathematical operation and report the answer to the correct number of significant figures? ( I have watche

professor190 [17]

0.008 ÷ 51.3 = 0.0002

Sig Figs

1

0.0002

Decimals

4

0.0002

Scientific Notation

2 × 10-4

E-Notation

2e-4

Words

zero point zero zero zero two

I HOPE I HELP

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3 years ago

Consider a 5 molecule system. Assume that the molecules can either be found on the right side or left side of their container.

8090 [49]

Answer:

The correct answer is 6 possible states

Explanation:

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3 years ago

The amount of matter in an object is called

marta [7]

Answer: Matter

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Matter is anything that has volume and/or mass.

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3 years ago