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OLga [1]
3 years ago
11

Which process is an example of a chemical change?

Chemistry
2 answers:
Alinara [238K]3 years ago
6 0

Answer:

d- activating a glow stick

Explanation:

Glow sticks produce light through a chemical reaction.

Ilia_Sergeevich [38]3 years ago
5 0

Answer:

burning of paper

melting of lead

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Let us calculate the structure of the electric shells of the Al atom. It has an atomic number of 13, so it has 13 electrons. The first 2 go to the first hell. The next 8 need to go to the second shell and the last 3 ones would go to the outermost shell. The outer shell, that is the most important one for chemical reactions, has thus 3 electrons. An atom always tries to have a completed outer shell (with either 2 or 8 atoms). It is easier for a cell to have a charge of +3 than a charge of -5 (smaller absolute value) and thus the Aluminum atom will try to get rid of the 3 electrons. In this process, it loses negative charge thus it will become positively charged. Hence, the correct answer is that it will prefer to lose 3 electrons and become positively charged.
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The day to day condition of atmosphere is known as
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d) atmosphere pressure

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True or false?<br><br> When sodium and chlorine combine and bond, a molecule is formed.
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When Sodium and Chlorine come together they transfer an electron.

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Hopefully this was clear and you understood!

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3 years ago
A short-lived structure formed during a collision is a(n)
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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
frez [133]

Answer : The mass of MnCO_3 required are, 35 kg

Explanation :

First we have to calculate the mass of MnO_2.

The first step balanced chemical reaction is:

2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2

Molar mass of MnCO_3 = 115 g/mole

Molar mass of MnO_2 = 87 g/mole

Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

The second step balanced chemical reaction is:

3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

Molar mass of MnO_2 = 87 g/mole

Molar mass of Mn = 55 g/mole

From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

4 0
3 years ago
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