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3 years ago

6

A marathon covers a distance of 26.2 miles. If 1 mile is equivalent to exactly 1760 yards, what is the distance of the race in y

ards? Use dimensional analysis.

1 answer:

Contact [7]3 years ago

8 0

Both mile and yard are units of distance thus they are inter convertible.

Distance covered in a marathon is 26.2 miles.

Since, 1 mile=1760 yards

Thus, distance can be converted into yards using the following conversion factor:

$\frac{1760 yards}{1 mile}$

Thus, distance covered in marathon in yard will be:

$d=26.2 mile(\frac{1760 yards}{1 mile})=46112 yards$

Therefore, distance of the race is 46112 yards.

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A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve

mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}$

= 0.0052 mol

Also, No. of moles = $\frac{mass}{\text{molar mass}}$

0.0052 mol = $\frac{mass}{2 g/mol}$

mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at $25^{o}C$

$P_{\text{water vapor}}$ = 24 mm Hg

= $\frac{24}{760}$ atm

= 0.03158 atm

Now, P = $\frac{756}{760} - 0.03158$

= 0.963 atm

Hence, n = $\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}$

= 0.0056 mol

So, mass of $H_{2}$ = 0.0056 mol × 2

= 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

= $\frac{0.01013 g}{0.0104 g} \times 100$

= 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

3 0

3 years ago

So guess what I did!!

Lapatulllka [165]

Answer:

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Explanation:

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7 0

3 years ago

Which of the following statements most likely

cupoosta [38]

Answer:

Forming a problem requires the scientist to use creativity to imagine new solutions.

Explanation:

Albert Einstein remains a critically prominent figure who conducted remarkable, ground-breaking research that not only formed the foundations of modern physics but also strongly affected the scientific world. It is difficult to teach imagination but it can be harnessed and accepted. Nothing incites our imaginative impulses we love more than the prospect of immediate creative inspiration. And creativity hits its full potential when paired with the experience, insights, and skills people gained by questioning the real-life problems.

6 0

3 years ago

Read 2 more answers

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

How many grams of solute are needed to prepare a 3.50% mass/mass solution that has a solution mass of 2.50x102 grams.

Elis [28]

One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.

Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute

5 0

3 years ago