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2 years ago

A rod measures 17.2 inches. How many centimeters is this? [ ? ] cm

Chemistry

1 answer:

Rufina [12.5K]2 years ago

3 0

This problem is providing the length in inches of a rod and it is being required to convert it to centimeters. According to the following, the answer will be 43.7 cm:

<h3>Units conversion:</h3><h3 />

In science, we use units conversion to represent measurements in different types of units depending on a specific requirement. To do so, we use conversion factors based on equivalence statements, relating the required unit with the given one.

In this case, the equivalence statement between inches and centimeters is 2.54 cm = 1 in and we can use it to get the answer as shown below:

$17.2in*\frac{2.54cm}{1in} \\\\=43.7 cm$

Learn more about units conversion: brainly.com/question/13016491

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Answer:

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3 years ago

Convert 26.02 x 1023 molecules of C2H8 to grams. Round your answer to the hundredths place.

aliina [53]

Answer:

x= 138.24 g

Explanation:

We use the avogradro's number

6.023 x 10^23 molecules -> 1 mol C2H8

26.02 x 10^23 molecules -> x

x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules

x= 4.32 mol C2H8

1 mol C2H8 -> 32 g

4.32 mol C2H8 -> x

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4 years ago

Petrochemicals are used in some cosmetics. True or False

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3 years ago

Which of the following best describes the particles that make up lightning and the particles that make up the water in a bottle?

Blizzard [7]

Answer:

Those in lightning are charged and those in water are not charged.

Explanation:

The particles in lightning differs from those in a water in that water molecules are neutral and those of lightning are charged.

Lightning is made up of charged ions or particles that have either lost or gained electrons.

Lightning is produced as a burst or discharge of charged particles that are imbalanced withing the clouds.

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3 years ago

Read 2 more answers

A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715

andrey2020 [161]

Answer:

$w_{Ar}=0.814$

Explanation:

Hello,

In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\ \\n=0.41mol$

Next, using the molar fraction of argon, we compute the moles of argon:

$n_{Ar}=0.41mol*0.715=0.29mol$

And the moles of methane:

$n_{CH_4}=0.41mol-0.29mol=0.12mol$

Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:

$w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol} }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}} \\\\w_{Ar}=0.814$

Regards.

7 0

3 years ago