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3 years ago

11

Which process is an example of a chemical change?

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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In a chemical reaction, the element 13al will most preferably ________. in a chemical reaction, the element 13al will most prefe

Rasek [7]

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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Answer:

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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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

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3 years ago

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