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2 years ago

7

If some of the gas bubbles from the reaction had escaped out the bottom of the tube, how would this have affected your value for

R?

1 answer:

Andrej [43]2 years ago

7 0

The composition would be more "diluted" in a sense.

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Water is composed of one atom of oxygen and two atoms of

egoroff_w [7]

Answer:

<em>Hydrogen.</em>

Explanation:

You've probably seen " $H_{2}0$ " which is the formula for water. It means that there's 2 hydrogen atoms, and one oxygen atom, in one molecule of water.

<em>Hope this helps! Feel free to mark me Brainliest if you feel this helped. :)</em>

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2 years ago

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2 years ago

In which of these situations is convection most likely the main form of heat transfer?

lorasvet [3.4K]

In the first situation: the mechanism of covection is the main form of heat transfer when warm air from a heater moves around and upward.

In the case of the metal pan the mechanism of heat transfer is conduction.

In the case of sunburn the mechanism is radiation.

In the case of an ice cube melting in a hand, conduction is the most important mechanism.

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2 years ago

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When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

<u>Answer:</u>

<u>For a:</u> The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

<u>For b:</u> The mass of nitric acid formed is 54.81 grams

<u>For c:</u> The mass of nitric acid formed is 206 grams

<u>Explanation:</u>

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

<u>For a:</u>

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

<u>For b:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

<u>For c:</u>

<u>For nitrogen dioxide:</u>

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

<u>For water:</u>

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

2 years ago

Which layer of soil often contains litter?

Lunna [17]

The O Layer. I believe. Or the A horizon

3 0

2 years ago