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DerKrebs [107]
2 years ago
5

A new computer chip contains 4.71×1022 Si atoms. Determine the mass (in grams) of the silicon present in the computer chip.

Chemistry
1 answer:
valina [46]2 years ago
8 0

The mass (in grams) of the silicon present in the computer chip is 2.19 g

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Si

But

1 mole of Si = 28 g

Thus, we can say that:

6.02×10²³ atoms = 28 g of Si

  • With the above information, we can obtain the mass of silicon, Si present in the computer chip. This can be obtained:

6.02×10²³ atoms = 28 g of Si

Therefore,

4.71×10²² atoms = (4.71×10²² × 28) / 6.02×10²³ g of Si

4.71×10²² atoms = 2.19 g of Si

Thus, the mass of silicon, Si present in the computer chip is 2.19 g

Learn more on Avogadro's number: brainly.com/question/25811549

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5 0
3 years ago
Read 2 more answers
If more solute can be dissolved in a solvent, the solution is
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Answer: B. Unsaturated

Explanation: in a saturated/super saturated solution, more solute will not be able to dissolve.

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]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<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 (II) bromide:</u>

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 %

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3 years ago
The chemical equation shown represents photosynthesis. Carbon dioxide plus A plus light with a right-pointing arrow towards B pl
astra-53 [7]

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3 0
2 years ago
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