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3 years ago

Arunner sprints 550 feet in 100 seconds. If the runner can hold this pace for an hour

Chemistry

1 answer:

Alex3 years ago

6 0

<h3>Answer:</h3>

Distance traveled is 9.055 km

<h3>Explanation:</h3>

Speed is given by dividing distance by time.

The distance given is 550 ft and the time is 100 seconds.

Therefore; Speed = 550 ft ÷ 100 sec

= 5.5 ft/sec

We can calculate the distance the runner covered in one and half hour.

Time = 1.5 hours

But 1 hour = 3600 sec

1.5 hours = (1.5 × 3600)

= 5400 sec

Distance is given by multiplying speed and time

Thus; Distance = 5.5 ft/sec × 5400 sec

= 29700 ft

But; 1 m = 3.28 ft

1000 m = 1 km

Therefore; 29700 ft will be equivalent to

= 29700 ft/3.28 ft

= 9054.878 M

but 1000 m = 1 km

Thus, distance = 9054.878 ÷ 1000 m

= 9.055 km

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Explanation:

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3 years ago

What is the density of a substance that weighs 340g and has a volume of 40cm3?

jeka94

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

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3 years ago

Is the sky a solid,liquid,or gas

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3 years ago

The pKb of the base cyclohexamine, C6H11NH2, is 3.36. What is the pKa of the conjugate acid, C6H11NH3

bonufazy [111]

Answer:

10.64

Explanation:

Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.

C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻ pKb = 3.36

C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:

C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq) pKa

We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.

pKa + pKb = 14.00

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4 years ago

How many grams are in 1.76 x 10^23 atoms of iodine

Mariana [72]

Answer:

$\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}$

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

$6.022*10^{23}$

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

$\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Multiply the given number of atoms by Avogadro's number.

$1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Flip the fraction so the atoms of iodine will cancel out.

$1.76*10^{23} \ atoms \ I*\frac{ 1 \ mol \ I}{6.022*10^{23} \ atoms \ I}$

$1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }$

Multiply so the problem condenses into 1 fraction.

$\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }$

$0.2922617071 \ mol \ I$

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

$\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply this fraction by the number of moles found above.

$0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply. The moles of iodine will cancel.

$0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }$

The 1 as a denominator is insignificant.

$0.2922617071 *{ 126.9045 \ g\ I }$

$37.08932581 \ g \ I$

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

$37.08932581 \ g \ I$

The 8 in the hundredth place tells us to round the 0 up to a 1.

$\approx 37.1\ g \ I$

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

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3 years ago