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alekssr [168]
4 years ago
14

Why is there no overall electrical charge on an aluminium atom?

Chemistry
1 answer:
Alex787 [66]4 years ago
3 0
The atoms always have an equal amount of protons and neutrons so lets say if one had 10 positive then it would have 10 negatives making it neutral.
10-10=0
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alexdok [17]

Answer:

He is percieved in a warm climate.

Explanation:

The frog doesn't feel too cold or too hot. He is just right.

5 0
3 years ago
What happens to the reactivity of nonmetals within a group from top of the group to the bottom?​
andreev551 [17]

As you go down a group, the atoms get bigger. The bigger the atom, the more unstable it is. The more unstable it is, the more reactive the atom will be. So, as you go down a group, the nonmetals will become more reactive.

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How can environmental science help you?​
alexandr402 [8]

Answer:

Organisms and humans depend on each other to get by. Environmental science is important because it enables you to understand how these relationships work.

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5 0
3 years ago
Use complete sentences to describe how making an ‘and’ compound sentence effects your solution set.
Lyrx [107]

Answer:

“And” indicates that both statements of the compound sentence are true at the same time. It is the overlap or intersection of the solution sets for the individual statements.

Explanation:

5 0
3 years ago
How many calories are required to convert 17 g of ice at 0.0°C to liquid water at 32.0°C? The heat of fusion of water is 80. c
Mazyrski [523]

Answer : The heat required is, 1904 calories.

Explanation :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(32.0^oC)

The expression used will be:

\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

m = mass of ice = 17 g

c_{p,l} = specific heat of liquid water = 1cal/g^oC

\Delta H_{fusion} = enthalpy change for fusion = 80.0cal/g

Now put all the given values in the above expression, we get:

\Delta H=17g\times 80.0cal/g+[17g\times 1cal/g^oC\times (32.0-0)^oC]

\Delta H=1904cal

Therefore, the heat required is, 1904 calories.

3 0
4 years ago
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