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QveST [7]
4 years ago
14

How many moles of magnesium ions are present in 283.8 mL of 6.74 M Mg3(PO4)2?

Chemistry
1 answer:
Dmitriy789 [7]4 years ago
6 0

Answer:1 mole of Mg3(PO4)2 (molecule) contains 3 moles of Mg atoms, 2 moles of P atoms and 2X4 moles of O atoms. = 0.03125 moles.

Explanation:

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Enzymes are biological catalysts which means they accelerate chemical reactions.
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4 0
3 years ago
How many molecules of Mg3N2 (magnesium nitride) are formed when excess Mg (magnesium)
dybincka [34]

Explanation:

<em>3Mg(s) + N2(g) = Mg3N2(s)</em>

First check that the equation is balanced. In this case, it is.

Assuming that magnesium is the limiting reactant:

  1. First find the molecular weight using the Periodic Table.

       We find that the atomic mass of magnesium is approximately

       <em>24.3g</em>, so the molecular weight is just <em>24.3g\mol</em>

   

    2. Next we need the mole to mole ratio. As there are <em>3</em>

        magnesiums for <em>1</em> magnesium nitride (shown by the coefficients), the                    

        mole to mole ratio is<em> 1 mol Mg3N2\3 mol Mg.</em>

   

    3. We need the amount of the substance, in grams. Since you have not    

        stated it in the question, I'll just do <em>10g</em> AS AN EXAMPLE. Note that    

       depending on the amount, the LIMITING REAGENT MAY DIFFER.

   4.  Finally, we need the molecular weight of <em>Mg3N2</em>, which we can easily    

        calculate to be around <em>100.9\mol.</em>

<em />

   5.  Putting this all together, we have<em> 10gMg⋅ (mol Mg\24.3gMg) </em>

<em>         (1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)</em>

     

        the units will cancel to leave <em>gMg3N2</em> (grams of magnesium nitride):

       

<em>        10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)</em>

<em>        (100.9g Mg3N2\mol Mg3N2)</em>

<em />

Doing the calculation yields approximately 13.84g.

Assuming that nitrogen is the limiting reactant:

Similarly, following the above steps but with <em>10g</em> of nitrogen yields <em>36.04g</em>

In conclusion, as we produce less amount of <em>Mg3N2</em> when we assumed that <em>Mg</em> was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

Note: This is in THIS CASE, where we have <em>10g</em> of both. The answer may vary depending on the amount of each substance.

7 0
3 years ago
Read 2 more answers
A sealed flask contains Ne, Ar, and Kr gas. If the total pressure in the flask is 3.782 atm, the partial pressure of Ne is 0.435
spin [16.1K]

Answer:

P_{Ar}=1.734atm = partial pressure of Ar

Explanation:

Acccoding to dalton's law of partial pressure total pressure of system is equal to alzebraic sum of all partial pressure.

P_{total} =P_{Ne}+P_{Ar} +P_{Kr}

P_{total}=3.782atm

P_{Ne}=0.435 atm

P_{Kr}=1.613atm

P_{Ar}=P_{total}-P_{Ne}-P_{Kr}

P_{Ar}=3.782-0.435-1.613

P_{Ar}=1.734atm =partial pressure of Ar

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4 years ago
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Henceforth, from left to right in the periodic table the tendency to gain electrons increases. In contrast, going down a group there is decreased association for electrons, because atomic radius increases which suggests that valence electrons are further away from the nucleus.This makes fluoride the ion with strongest association of electrons. The noble gases have a complete shell so cannot attract electrons to themselves, which means they have no electronegativity.

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3 years ago
He specific rate constant, k, for the following first-order
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Hope this helps. Thanks

5 0
3 years ago
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