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3 years ago

What is the mechanism by which Ozone is formed in the presence of sunshine?

1 answer:

8 0

Stratospheric ozone is formed naturally through the interaction of UV radiation with molecular oxygen.

not sure if this is what you want but hope it helps!!!

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Choose the answer that best completes the following statement: When an aluminum atom reacts so as to attain a noble gas electron

boyakko [2]

The options

Select one:

a. a 3- ion forms.

b. the noble gas configuration of argon is achieved.

c. the result is a configuration of 1s2 2s2 2p6.

d. the atom gains five electrons.

Answer:

c. the result is a configuration of 1s2 2s2 2p6.

Explanation:

Aluminium atom has atomic number of 13 , hence the number of electron is 13 for a neutral atom of aluminium. When aluminium atom reacts with other elements it usually gives out three electron to attain the octet configuration.

The cation representation of aluminium is Al3+ because it has loss three electron to attain the octet rule. Aluminium will be left with 10 electrons after losing 3 of it electrons. The electronic configuration will be represented as follows after losing three electrons;

1S² 2S² 2P∧6 .

At this stage the octet rule has been achieved as it will be represented as

2 8. The first energy shell now contains two electron and the second energy shell contains 8 electrons.

The configuration of Neon has been formed in the process.

4 0

3 years ago

What is the balanced equation for naphthalene and sulphur?

Marianna [84]

napthalene: C10H8 sulphur: S02

Explanation:

C10H8

7 0

3 years ago

What quantity is measured in mol/dm³?

natima [27]

L
$M = \frac{n}{V} \\ M = \frac{mol}{L} \\ or \\ M = \frac{mol}{ {dm}^{3} }$
mol/dm³ is measure for molarity

3 0

3 years ago

Convert 0.0052 meters to millimeters

amm1812

Answer:

5.2 millimeters

Explanation:

hope this helps :)

5 0

3 years ago

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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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