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2 years ago

The molar mass of silicon (Si) is 28.09 g/mol. Calculate the number of atoms in a 92.8 mg sample of Si.

Chemistry

1 answer:

adelina 88 [10]2 years ago

6 0

Explanation:

one mole of any substance there are 6.022×1023 units of that substance. (This number is called Avogadro's number, NA.)

We need to convert the mass of silicon to moles using the molar mass of silicon, 28.06gmol. This number means that one mole of pure silicon would have a mass of 28.06g. Our given mass, however, is in milligrams; to convert this to grams we'll use the conversion factor 1g103mg:

5.86mg Si(1g103mg)=0.00586g Si

Now, using silicon's molar mass, we'll convert this mass to moles of Si:

0.00586g Si(1mol Si28.06g Si)=2.09×10−4mol Si

Finally, let's use Avogadro's number to convert

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Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l)

Marina CMI [18]

Answer:

Please see the complete formt of the question below

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)

According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.

The answer to the above question is

The limiting reactant is the MnO₂

Explanation:

To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂

Molar mass of MnO₂ = 86.9368 g/mol

Molar mass of HCl = 36.46 g/mol

From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂

However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed

7 0

2 years ago

2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta

Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

$\eta = \frac{m}{M}$

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

$\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles$

$\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles$

$\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles$

$\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles$

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

$\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles$

We have that between MnO₂ and KOH, the limiting reactant is KOH.

$\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles$

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

$\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles$

Therefore, the limiting reactant is KOH.

I hope it helps you!

6 0

2 years ago

Read 2 more answers

What happens during a mosquito 6 chromosomes to form eggs.

lbvjy [14]

<span>Since the diploid number of mosquitoes is six, that means that there are 6 chromosomes in every somatic (non-reproductive) cell,including the cells that make up the stomach.The six chromosomes represent two three-chromosome sets the mosquito received one set of three chromosomes from the egg cell of the mother and a similar set of three chromosomes from the sperm cell of the father.</span>

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2 years ago

How many moles are there in 3.612 X 10^24 molecules of phosgene (COCL2)

Ugo [173]

Answer:

Explanation:

No of molecules = 3.612*10(24)

Avogadro's number = 6.022*10(23)

no of moles = No of molecules/Avogadro's number

No of moles =3.612*10(24)/6.022*10(23)

No of moles = 6mol

3 0

2 years ago

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In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 27.0 ∘C . You identified that the ideal pressure

dimaraw [331]

Answer : The percent difference between the ideal and real gas is, 4.06 %.

Explanation : Given,

Ideal pressure (true value) = 49.3 atm

Real pressure (measured value) = 47.3 atm

The formula used to calculate percent difference is :

Percent difference = $\frac{\text{True value - Measured value}}{\text{True value}} \times 100$

Percent difference = $\frac{(49.3- 47.3)atm}{49.3atm}\times 100$

Percent difference = 4.06 %

Therefore, the percent difference between the ideal and real gas is, 4.06 %.

4 0

2 years ago