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3 years ago

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What is the most common of all compounds are composed of molecules whose atoms are arranged in a straight chain, a branched chai

n, or a ring

1 answer:

Gemiola [76]3 years ago

3 0

<span>Sugar's carbon molecules can be arranged in a straight chain, a branched chain, or a ring.</span>

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Which list of phases of H2O is arranged in order of increasing entropy

vladimir2022 [97]

Answer: Solid, liquid, and gas are the three phases of H2O that are listed in order of increasing entropy.

The haphazard motion of water molecules in molecular water gives it the maximum flexibility.

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2 years ago

An aqueous solution containing 7.96 g7.96 g of lead(II) nitrate is added to an aqueous solution containing 6.82 g6.82 g of potas

Eduardwww [97]

<u>Answer:</u> Lead nitrate is the limiting reagent and potassium chloride is the excess reagent. the amount of excess reagent left is 3.205 grams. The amount of precipitate (lead chloride) recovered is 5.96 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For lead nitrate:</u>

Given mass of lead nitrate = 7.96 g

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

$\text{Moles of lead nitrate}=\frac{7.96g}{331.2g/mol}=0.024mol$

<u>For potassium chloride:</u>

Given mass of potassium chloride = 6.82 g

Molar mass of potassium chloride = 74.55 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium chloride}=\frac{6.82g}{74.55g/mol}=0.091mol$

For the given chemical equation:

$Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)$

By Stoichiometry of the reaction:

1 mole of lead nitrate reacts with 2 moles of potassium chloride.

So, 0.024 moles of lead nitrate will react with = $\frac{1}{2}\times 0.024=0.048moles$ of potassium chloride

As, given amount of potassium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, lead nitrate is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (potassium chloride) left = 0.091 - 0.048 = 0.043 moles

Now, calculating the mass of potassium chloride from equation 1, we get:

Given mass of potassium chloride = 6.82 g

Molar mass of potassium chloride = 74.55 g/mol

Putting values in above equation, we get:

$0.043mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=3.205g$

By Stoichiometry of the reaction:

1 mole of lead nitrate produces 1 mole of lead chloride.

So, 0.024 moles of lead nitrate will produce = $\frac{1}{1}\times 0.024=0.024moles$ of lead chloride

Now, calculating the mass of lead chloride from equation 1, we get:

Molar mass of lead chloride = 278.1 g/mol

Moles of lead chloride = 0.024 moles

Putting values in equation 1, we get:

$0.024mol=\frac{\text{Mass of lead chloride}}{278.1g/mol}\\\\\text{Mass of lead chloride}=6.6744g$

We are given:

Percentage yield of the reaction = 89.3 %

So, amount of lead chloride recovered will be = $\frac{89.3}{100}\times 6.6744g=5.96g$

Hence, lead nitrate is the limiting reagent and potassium chloride is the excess reagent. the amount of excess reagent left is 3.205 grams. The amount of precipitate (lead chloride) recovered is 5.96 g.

4 0

3 years ago

Which of the following points is/are true for plasma?

Likurg_2 [28]

Answer:

4 all of the choices. the rest of the answers are correct so it's all of the choices

5 0

2 years ago

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The diagram shows heat energy being transferred as hot water rises to the top of the pot and cool water sinks to the bottom.

Orlov [11]

I think it would be insulation

7 0

3 years ago

2p is the correct representation for the sub-shell with n = 2 and l = 1.

boyakko [2]

Explanation:

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

s = 1 orbital

p = 3 orbitals

d = 5 orbitals

f = 7 orbitals

For n = 4

l = 0 to (n-1) = 0 to 3 = (4s , 4p , 4d , 4f)

Number of subshells = 4

Number of orbitals = 1 + 3 + 5 + 7 = 16

The maximum number of electrons the n = 4 shell can contain:

Each orbital can holds upto two electrons, then 16 orbitals will have :

$16\times 2=32$

32 is the maximum number of electrons the n = 4 shell can contain

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3 years ago

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