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otez555 [7]
3 years ago
5

A car traveling at a constant speed travels 175 miles in four hours. How many minutes will it take for the car to travel 1000 fe

et? (5,280 feet=1 mile)
Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

It takes 0.26 minutes to travel 1000 feet

<u>Solution:</u>

Given that car traveling at a constant speed travels 175 miles in four hours

Distance = 175 miles

Time taken = 4 hours

Convert the units to feet and minute

Given that,

1 mile = 5280 feet

175 miles = 5280 x 175 feet = 918720 feet

Also we know that,

1 hour = 60 minutes

4 hours = 60 x 4 minutes = 240 minutes

Thus, we got,

Distance = 918720 feet

Time taken = 240 minutes

<em><u>Find the speed of car</u></em>

<em><u>Speed is given by formula:</u></em>

speed = \frac{distance}{time}

speed = \frac{918720}{240} = 3828

Thus speed of car is 3828 feet per minute

<em><u>How many minutes will it take for the car to travel 1000 feet?</u></em>

Let "x" be the minutes needed for 1000 feet

speed of car is 3828 feet per minute

1 minute = 3828 feet

x minute = 1000 feet

This forms a proportion. We can solve by cross multiplying

x \times 3828 = 1000 \times 1\\\\x = \frac{1000}{3828}\\\\x = 0.26

Thus it takes 0.26 minutes to travel 1000 feet

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Manganese is a transition metal. Consider the isotope: Mn-53. How many protons are in an atom of Mn-53 if the atom has a charge
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Answer:

25

Explanation:

Isotopes of an element are the same element with different mass numbers but the same atomic number or the number of protons.

we know that in a balanced atom

Number of electrons = Number of protons = Atomic number

Atomic number (z) of manganese = 25

For Mn+5, the number of protons remains the same = 25

However, the number of electrons on Mn+5 = 21

7 0
3 years ago
The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B
Alborosie

<u>Answer:</u> The empirical formula for the given compound is C_{15}H_{24}O_1

<u>Explanation:</u>

The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=5.995g

Mass of H_2O=1.963g

Mass of sample = 2.001 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

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

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

So, in 5.995 g of carbon dioxide, \frac{12}{44}\times 5.995=1.635g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 1.963 g of water, \frac{2}{18}\times 1.963=0.218g of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

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

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

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles

  • <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 0.0092 moles.

For Carbon = \frac{0.136}{0.0092}=14.78\approx 15

For Hydrogen = \frac{0.218}{0.0092}=23.69\approx 24

For Oxygen = \frac{0.0092}{0.0092}=1

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

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is C_{15}H_{24}O_1

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Answer:

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Explanation:

That's the definition hope it helps.

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