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2 years ago

7

What happens to the air as it gets farther from the heat source?

1 answer:

Effectus [21]2 years ago

6 0

Answer:

molecules heat and move faster, they are moving apart. So air, like most other substances, expands when heated and contracts when cooled.

Explanation:

Because there is more space between the molecules, the air is less dense than the surrounding matter and the hot air floats upward.

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Martian periodic table

denis23 [38]

<span>Chemists considered Martian Periodic Table as one of the most useful tools they’ve used whose purpose is to arrange the Martian elements according to their properties and their atomic number. On this periodic table, it was found out that the properties seem to vary. Dmitri Mendeleev, the first to arrange the periodic table according to atomic mass in a manner that the elements with the same properties were grouped together. It was Henry Moseley who later arranged the periodic table in accord to the increasing number of atoms per element. </span>

3 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

2 years ago

Tomato juice has a pH of 4.20. Calculate the [H3O+] and [OH–] in tomato juice.

Free_Kalibri [48]

<span>[H3O+] = 10^(-pH) = 10^(-4.20) = 6.3 x 10^-5 M

pOH = 14 - pH = 14 - 4.20 = 9.80

[OH-] = 10^(-pOH) = 10^(-9.80) = 1.6 x 10^-10 M
</span>

8 0

2 years ago

How many atoms are in 0.00014 mol of Na?

Ludmilka [50]

Answer: Use Avogadro's number, NA = 6.02 × 1023 atoms/mol.

Explanation:

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2 years ago

What are the two parts of a solution?

Vlad [161]

The 2 parts or components that make up a solution would be the solute and the solvent.

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2 years ago