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3 years ago

The repeating subunits that are responsible for the chance of a crystal are know as

Chemistry

1 answer:

Naddik [55]3 years ago

7 0

Unit cells.
They are the smallest group of atoms that take form of a crystal and can repeat in 3 dimensions.

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why is it much easier for group 14 elements to become stable by sharing instead of transferring electrons9

Effectus [21]

Because they are closer to the farther end of the periodic table. Since they are closer to the farther end they don't want to give away their electrons because it would be easier for them to just steal them from other atoms.

5 0

3 years ago

Read 2 more answers

I’m not sure quite sure about this one please help

ehidna [41]

This one is correct

3 0

3 years ago

In the combined gas law, if the volume is decreased, what happens to the temperature if pressure remains constant?

Flura [38]

The correct answer is A. In the combined gas law, if the volume is decreased and the pressure is constant, then the temperature decreases.

P1V1/ T1 = P2V2 / T2

Assume the volume decrease by half; V2 = V1/2

P1V1/ T1 = P2V1 /2 T2

Cancelling terms,

1/T1 = 1/2 T2

T2 = T1/2

Thus, the temperature decreased.

8 0

3 years ago

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

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

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

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

I need the answer for this question

ch4aika [34]

I want to say A only... Hope I helped!!

5 0

3 years ago