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frozen [14]
1 year ago
7

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]1 year 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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The greatest amount of energy released per gram of reactants occurs during a (1) redox reaction, although it should be noted that there are exceptions depending on environment. 
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1 year ago
A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2
4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

C_3H_7+5O_2\rightarrow 3CO_2+4H_2O

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

\frac{3}{1}\times 3.85 mol=11.55 mol of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

\frac{4}{1}\times 3.85 mol=15.4 mol of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

\frac{4}{5}\times 0.647 mol=0.5176 mol of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

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\frac{1}{5}\times 0.647 mol=0.1294 mol of propane.

0.1294 moles of propane are consumed.

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Is the aqueous solution of each of these salts acidic, basic, or neutral? (a) cr(no3)3 acidic basic neutral (b) nahs acidic basi
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1 year ago
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A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
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Answer:

M_{Na^+}=1.36M

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Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

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Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

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<span>RE: "numerically to the hundredths place." </span>

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4 0
2 years ago
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