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german
3 years ago
6

How do the strong ionic bonds affect the boiling and melting points of salts

Chemistry
1 answer:
likoan [24]3 years ago
3 0

Answer:

The more energy needed, the higher the melting point or boiling point . Since the electrostatic forces of attraction between oppositely charged ions are strong, their melting and boiling points are high.

Explanation:

hope this helps.

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A solid is held in shape by strong forces.
n200080 [17]

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What is the full meaning of DR.HERC.​
n200080 [17]

Answer:

It is

D- <u><em>DEFINE</em></u> the problem

R-<em><u>RESEARCH</u></em> on the problem

H- Carry out a <u><em>HYPOTHESIS</em></u>

E- carry out an <em><u>EXPERIMENT</u></em>.

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C-summarise the <u><em>CONCLUSION</em></u>.

Explanation:

Hope it helps.

5 0
3 years ago
Is 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O a double replacement?
natulia [17]

Answer:

The answer to your question is No, it is not.

Explanation:

Data

                C₄H₁₀  +  13O₂   ⇒   8CO₂ +  10H₂O

In a double replacement reaction, two reactants interchange cations an example of these reactions are neutralization reactions. In neutralization reactions, an acid and a base react to form a salt and water.

The reaction of this problem is not a double replacement reaction because the products are carbon dioxide and water, not a salt and water.

The reaction of this problem is a combustion reaction.

8 0
4 years ago
A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this r
EastWind [94]
M_{Al}=26,98\frac{g}{mol}\\&#10;m=51g\\\\&#10;n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\&#10;M_{NaOH}=39,4\frac{g}{mol}\\&#10;m=84,1g\\\\&#10;n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span>  2mol      :     2mol           :                                  3mol
2,14mol   :     1,89mol      :                                  2,835mol
remains         completely consumed
2,14-1,89=0,25mol


A) Al

B)  

M_{NaOH}=39,4\frac{g}{mol}\\&#10;n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g

C)
n=2,835mol\\&#10;M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g

7 0
3 years ago
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