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max2010maxim [7]
2 years ago
14

A 13 Ohm resistor, R, is connected to a battery with unknown voltage, V, as shown below. The total current, I, flowing through t

he circuit is 1.24 Amperes. Find the unknown voltage, V, of the battery in Volts. Round your answer to 2 decimal places
Chemistry
1 answer:
wlad13 [49]2 years ago
3 0

Answer:

16.12 V

Explanation:

Ohms law explains or portrays the relationship between the voltage (V), current (I) and resistance (R). This is represented using the equation as follows:

V = IR

Where;

V = voltage (Volts)

I = current (amperes)

R = resistance (ohms)

According to the information provided in the question, resistance (R) = 13 Ohm, current (I) = 1.24 Amperes, V = ?

V = 13 × 1.24

V = 16.12 V

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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

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It is important for scientists to have a standard system of naming compounds and other chemical procedures to avoid ambiguity in the nomenclature of chemical compounds and procedures.

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Therefore, it is imperative that a unified and standard system of nomenclature should be devised so that all scientists can have the same name for compounds. This reduces ambiguity associated with the nomenclature of chemical substances and procedure.

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