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3 years ago

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The speed of light in a vacuum is 2.998 × 10 8 8 m/s. How long does it take for light to circumnavigate the Earth, which has a c

ircumference of around 24,900 miles? Do not enter “seconds” as part of your answer.

1 answer:

svp [43]3 years ago

8 0

Answer:- 0.134 seconds

Solution:- The speed is given as $2.988*10^8\frac{m}{s}$ and the circumference is 24900 miles which is same as the distance light have to covered. It asks to calculate the time required to cover this distance by the light.

We need to do unit conversion from miles to meter as the speed is given in meter per second.

1 mile = 1609.34 meter

So, $24900mile(\frac{1609.34meter}{mile})$

= 40072566 meters

Know that, $speed=\frac{distance}{time}$

It's rearranged to time as, $time=\frac{distance}{speed}$

Let's plug in the values in it:

$time=\frac{40072566meter}{2.988*10^8\frac{meter}{second}}$

= 0.134 seconds

So, the light would take 0.134 seconds to travel the mentioned speed. The answer without the unit is 0.134.

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The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is $CaCN_2$

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

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