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1 year ago

Briefly account for the following relative values:

Chemistry

1 answer:

Sedbober [7]1 year ago

4 0

The melting points of Na and K are 89°C and 63°C, respectively because of weak attractive force.

Sodium and potassium belongs to group 1.

<h3>What is atomic radii? </h3>

The distance of the outermost shell to the nucleus is called atomic radii.

As we go down the group, new shell is added which means number of shell increase as we go down the group.

Due to increase in the number of shells, the distance of outermost shell to the positively charged nucleus increase.Due to which atomic radius increase.

Due to increase in the atomic radii, the nuclear force of attraction on outermost shell decreases down the group.

This decrease in the nuclear force of attraction cause decrease in the melting point.

We can say that, melting point of a compound is inversely proportional to the atomic radii.

Therefore, melting point of sodium is less than the potassium atom.

Thus, we concluded that the melting points of Na and K are 89°C and 63°C, respectively because of weak attractive force.

learn more about melting point:

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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.

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Answer:

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Explanation:

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We can say that:

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

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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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