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Veronika [31]
3 years ago
6

Using Avogadro's Number (6.02*10^23): Calculate the number of molecules in 3.00 moles H2S . Express your answer numerically in m

olecules.. Calculate the number of moles of Cl atoms in 1.81×10^24 formula units of magnesium chloride,MgCl2 .
Chemistry
2 answers:
gulaghasi [49]3 years ago
6 0

Answer:

1.81x10^2^4~molecules~of~H_2S

3~mol~Cl

Explanation:

For this question we have to take into account that the<u> ratio between moles and molecules</u>:

1~mol~H_2S~=~6.023x10^2^3~molecules~of~H_2S

Then, using this ratio we can do the <u>conversion:</u>

3.00~mol~H_2S\frac{6.023x10^2^3~molecules~of~H_2S}{1~mol~H_2S}

1.81x10^2^4~molecules~of~H_2S

For the next question we have to keep in mind that the formula units are a <u>synonym</u> for molecules and we have to use<u> two ratios</u>:

1~mol~MgCl_2=2~mol~Cl

1~mol~MgCl_2~=~6.023x10^2^3~molecules~of~MgCl_2

<u>Using these two ratios</u>:

1.81x10^2^4~molecules~MgCl_2\frac{1~mol~MgCl_2}{6.023x10^2^3~molecules~of~MgCl_2}\frac{2~mol~C}{1~mol~MgCl_2}

3~mol~Cl~

mars1129 [50]3 years ago
3 0
Avagadro's number is just a measurement. One mole is 6.022 X 10^23 of anything - atoms, molecules, marbles... anything. 
<span>1) If one mole = 6.022 X 10^23, then 8.00mol of H2S is: </span>
<span>(3.00mol H2S) (6.022 X 10^23 molecules H2S / 1 mol H2S) = 1.8060 X 10^24 molecules H2S. </span>
<span>Rounded to 3 sig figs =1.81 X 10^24 molecules H2S 
</span>part2.
<span> This one uses moles in the stoichiometric sense as well as the measurement. One formula unit of MgCl2 contains 1 mole Mg and 2 moles Cl. </span>
<span>First, figure out how many moles of formula units there are. </span>
(1.81 X 10^24 FU's) (1mol MgCl2 / 6.022 X 10^23 FU's) = 3.0056mol MgCl2.

<span>Now, we know that there are 2 moles of Cl in every mole of MgCl2 (2 Cl atoms in every unit of MgCl2). From this we can determine how many moles of Cl atoms there are: </span>
<span>(3.0056mol MgCl2) (2mol Cl atoms / 1mol MgCl2) = 6.0112mol Cl atoms. </span>
<span>Now round to 3 sig figs = 10.0mol Cl atoms</span>
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Answer:

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A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and
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Answer:

5

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

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What mass in grams would 5.7L of hydrogen gas occupy at STP?​
tekilochka [14]

Answer:  The correct answer is:  " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:  

H₂ (g)  ;   that is:  a "diatomic element" (diatomic gas) ;

_________________________________________

The molecular weight of "H" is:  1.00794 g ;

   (From the Periodic Table of Elements).

So, the molecular weight of:  H₂ (g)  is:

    " 1.00794 g * 2  = 2.01588 g ; {use calculator) ;

_________________________________________

Note the conversion for a gas at STP:

______

  1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So:     " 5.7 L H₂ (g)  *  \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;

The "L" ("literes" cancel out to "1" ;  since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

 [5.7 * 2.104588 g ] / 22.4  =  ?  g ;

______________________

→ [ 11.9961516  g ] / 22.4 =

          0.53554248214  g ;l

_____________________________

We round this value to:  " 0.54 g " ;

 → since "5.7 L " has 2 (two) significant figures;  

     22.4 is an exact number conversion;

     and "5.7 L" has fewer significant figures than:

    " 2.104588 " ; or:  " 1.00794 " .

  → as such: We round to "2 (two) significant figures."

______________________________

Hope this is helpful.  Wishing you the best in your academic endeavors!

_______________________________

8 0
3 years ago
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