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3 years ago

Which of the following measurements is expressed to three significant figures? 7.30 × 10–7 km 0.070 mm 0.007 m 7077 mg

Chemistry

2 answers:

Aneli [31]3 years ago

5 0

Answer : The correct option is, $7.30\times 10^{-7}Km$

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures are :

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

$7.30\times 10^{-7}Km$ has 3 significant figures.

0.007 has 1 significant figure.

0.070 has 2 significant figures.

7077 has infinite significant figures.

Hence, the correct option is, $7.30\times 10^{-7}Km$

trapecia [35]3 years ago

4 0

7.30 x 10-7

Good luck.

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Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

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V = Volume of the gas = 2.19 L

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To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

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