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djverab [1.8K]
2 years ago
12

How many molecules are in 4.5 moles of CO2?

Chemistry
1 answer:
serg [7]2 years ago
5 0

Answer:

2.71 × 10^24 molecules

Explanation:

We are given;

Moles of CO₂ as 4.5 moles

We are required to determine the number of molecules

From the Avogadro's constant;

1 mole of a molecular compound contains 6.022 × 10^23 molecules

Therefore;

Molecules of CO₂ = Moles of CO₂ × Avogadro's constant

That is;  

               = 4.5 moles × 6.022 × 10^23 molecules/mole

               = 2.71 × 10^24 molecules

Hence, 4.5 moles of CO₂ contains 2.71 × 10^24 molecules

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A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
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Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




7 0
3 years ago
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