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PolarNik [594]
3 years ago
6

7. An alpha particle is_______

Chemistry
1 answer:
Rainbow [258]3 years ago
4 0

Answer:

b. positively

Explanation:

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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
If the density of gold is 19.3 g/cm3 what would be the volume of 550 g of gold?​
RUDIKE [14]

Answer:

28.497 cm3

Explanation:

Formula

D=m/v

Given data:

density = 19.3g/cm3

mass = 550 g

Now we will put the values in formula:

V=m/d

V=550 g/ 19.3 g/cm3  = 28.497 cm3

So the volume of gold is 28.497 cm3.

7 0
3 years ago
Plz help!!!!!! will mark as brainliest!!!!!!!!
maksim [4K]
A.a low basket with plastic liner
7 0
3 years ago
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Hiii please help me to balance the chemical equation:
Fed [463]

Answer:

2Na + 2H2O = 2NaOH + H2

Explanation:

hope this helps

4 0
3 years ago
If the hydrogen ion concentration is 10-7M, what is the pH of the solution?<br><br> Show all work.
Kobotan [32]
Remember pH=-log(H ions). So it would be pH=-log(10^-7).
6 0
4 years ago
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