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Tpy6a [65]
3 years ago
8

Each element has a unique number of ____, called the atomic number. since an atom is neutral this number is also equal to the nu

mber of _____ present
Chemistry
1 answer:
Tju [1.3M]3 years ago
8 0
1.) protons
2.)electrons
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What quantity of potassium phosphate, in grams, is required to prepare 500.0 ml of solution where you the concentration of potas
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Answer: 3.54

Explanation:

You're forgetting to divide by 3 for the 3 moles of potassium that are in potassium phosphate.

Potassium Phosphate= K3PO4

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Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling
Reil [10]

Answer:

The boiling  point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>

Explanation:

The formula for molal boiling Point elevation is :

\Delta T_{b} = iK_{b}m

\Delta T_{b} = elevation in boiling Point

K_{b} = Boiling point constant( ebullioscopic constant)

m = molality of the solution

<em>i =</em> Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

<em>i =</em> Van't Hoff Factor = 5

m = 8.5 m

K_{b} = 0.512 °C/m

Insert the values and calculate temperature change:

\Delta T_{b} = iK_{b}m

\Delta T_{b} = 5\times 0.512\times 8.5

\Delta T_{b} = 21.76 K

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

\Delta T_{b} = T_{b} - T_{b}_{pure}

T_{b}_{pure} = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0
3 years ago
Can someone help me here please i need this :(​
podryga [215]
Start by adding the numbers then divide
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