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Tema [17]
3 years ago
6

When a neutral atom loses an electron what happens to its size

Chemistry
1 answer:
emmainna [20.7K]3 years ago
4 0
Its outer shell will be removed thus having a smaller iconic radius.
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Solve using significant figures <br> 8.647 + 45.969
Murrr4er [49]

Answer:

54.616

Explanation:

8.647 + 45.969

or rewrite for easier look:

45.969 +

 8.647 =

54.616

Hope this helped :3

 

5 0
4 years ago
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Which form of emission is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass
oksano4ka [1.4K]

Answer: It is Gamma ray

4 0
3 years ago
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How much heat does it take to melt 2.5 kg of lead when it is at its melting point (m.p. = 327.5°C; Hf = 2.04 × 104 J/kg)?
irinina [24]
1. the amount of heat that it would take would be :
5.1 x 10^4 J

2. I think the substance would be : Iron

3. the amount of heat required would be :
1.13 x 10^4 kJ

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6 0
3 years ago
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Help plz ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!
ddd [48]
Honestly I don’t even know
7 0
3 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
3 years ago
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