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3 years ago

These models show the electron structures of two different nonmetal elements.

Chemistry

1 answer:

Soloha48 [4]3 years ago

4 0

Answer: Option (D) is the correct answer.

Explanation:

Valence shell is the shell present on the outermost core of an atom and electrons present in the valence shell are known as valence electrons.

If an atom has completely filled valence shell then it means the atom is not reactive in nature because it is already stable.

But when an atom has less than eight electrons in its valence shell then it means to attain stability the atom will readily attract electrons towards itself.

As the given element 1 has 8 electrons in its valence shell. Hence, it is not reactive in nature but element 2 has 6 valence electrons. So, in order to attain stability element 2 will readily attract 2 electrons from a donor atom.

Thus, we can conclude that element 2 is more reactive because it does not have a full valence shell, so it will attract electrons.

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

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3 years ago

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Marina86 [1]

Answer:

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3 years ago

In a chemical reaction between copper metal and silver nitrate, 12.7 g of copper(II) and 64.5 g of silver nitrate react, and 38.

hodyreva [135]

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3 years ago

What is the oxidation state of Hg in Hg2Cl2?

Anna [14]

Answer:

Explanation:

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Paha777 [63]

I think C, but I could be wrong

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