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Leno4ka [110]
4 years ago
6

Magnesium phosphate is used as a dental polishing agent. how many moles of ions (cations and anions together) are in 1 mole of m

agnesium phosphate, mg3(po4)2?
Chemistry
1 answer:
S_A_V [24]4 years ago
7 0
The answer to this question would be:5 mol

To answer this question, you need to know the chemical equation of mg3(po4)2 when forming an ion. The equation should be: <span>mg3(po4)2= 3Mg(2+)  +2PO4(3-)

That means for 1 mol of </span><span>mg3(po4)2 there will be 3 magnesium(cation) ion and 2 phosphates(anion) ion, the result would be 5 mol of ion.</span>
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Distillation is a process of separating the component substances from a liquid mixture by selective evaporation and condensation. 
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A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21 c and the pressure is lowered to 0.80 atm,
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Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

\frac{P*V}{T}=k

Having an initial state 1 and a final state 2 it is possible to say that:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

  • P1= 1 atm
  • V1= 1.2 L
  • T1= 273 °K
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  • V2= ?
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Replacing:

\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}

Solving:

V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}

V2= 1.62 L

<u><em>The new volume is 1.62 L</em></u>

<u><em></em></u>

8 0
3 years ago
Enter the oxidation number of one atom of each element in each reactant and product.
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Answer:

1. -4

2. +1

3. 0

4. +4

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6. +1

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Explanation:

Know oxidation rules.

- Hope this helped! Please let me know if you would like to learn this. I could show you the rules and help you work through them.

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