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Murrr4er [49]
3 years ago
10

The speed of light is around 6.706×10^8 miles per hour. What is the speed of light in units of miles per minute?

Chemistry
1 answer:
Sindrei [870]3 years ago
7 0

<u>Answer:</u> The speed of light in miles per minutes is 1.117\times 10^7miles/min

<u>Explanation:</u>

We are given the speed of light is 6.706\times 10^8miles/hr and we need to convert it into miles/min. So, we use the converion factor:

1 hour = 60 minutes

Converting that quantity into miles/minutes, we get:

(\frac{6.706miles}{1hr})(\frac{1hr}{60min})=\frac{1.117\times 10^7miles}{min}

Hence, the speed of light in miles per minutes is 1.117\times 10^7miles/min

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Svet_ta [14]

Answer:

the  number of laps in the case when he run for 50 minutes is 18,333.33

Explanation:

The computation of the number of laps in the case when he run for 50 minutes is shown below:

Given that

He runs 440m lap in 1.2 minutes

So in 50 minutes he can have laps of

= 440 × 50 ÷ 1.2

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Answer:

a)

Fe^{2+}⇒Fe^{3+}+e^-

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b)

Mg⇒Mg^{2+}+2e^-

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Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

Fe^{2+}⇒Fe^{3+}+e^-

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

Br_2+2e^-⇒2Br^-

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

Mg⇒Mg^{2+}+2e^-

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

Cr^{3+}+e^-⇒Cr^{3+}

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