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Law Incorporation [45]
3 years ago
10

Using the trends in the periodic table, rank the following atoms in order of increasing electronegativity

Chemistry
1 answer:
Kay [80]3 years ago
5 0
F >O>N>…………… 2nd row
in the third row is the same

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HELPPPPPP PLZZZZZ! FILE ATTACHED BELOW! ASAP PLZZZ!
baherus [9]
90 would be my guess.

4 0
3 years ago
A reaction that had two compounds as reactants and two compounds as products is most likely a
sergejj [24]

double-replacement reaction

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3 years ago
What differs from between a radioactive isotope and a stable isotope
Digiron [165]

A stable isotope has just<em> the right number of neutrons for the number of protons </em>(the <em>n:p ratio</em>) to hold the nucleus together against the repulsions of the protons.

A radioactive isotope has either too few or too many neutrons for the nucleus to be stable,

The nucleus will then emit <em>alpha, beta, or gamma radiation</em> in an attempt to become more stable.  

7 0
4 years ago
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Kryger [21]

Answer:Bbobobob Kittie

Explanation:

5 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
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