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3 years ago

The Average Speed of the orbiting space shuttle is

Chemistry

1 answer:

maxonik [38]3 years ago

8 0

Explanation:

It is given that, the Average Speed of the orbiting space shuttle is 17500 miles/hour.

We need to convert the speed in kilometers/ second

We know that,

1 mile = 1.609 km

1 km = 0.621 miles

1 hour = 3600 seconds

$17500\ \dfrac{\text{miles}}{\text{hour}}=17500\ \dfrac{\text{miles}}{\text{h}}\times \dfrac{1\ h}{3600\ s}\\\\=17500\times \dfrac{\text{miles}}{3600\ s}$

Now cancel the miles in numerator.

$17500\times \dfrac{\text{miles}}{3600\ s}=17500\times \dfrac{\text{miles}}{3600\ s}\times \dfrac{1.609\ km}{1\ \text{miles}}\\\\=17500\times \dfrac{1.609}{3600}\ km/s\\\\=7.82\ km/s$

So, 17500 miles/hour is equal to 7.82 km/s.

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A student was synthesizing aspirin in the laboratory. Using the amount of limiting reactant, she

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A biochemist carefully measures the molarity of magnesium ion in of cell growth medium to be . Unfortunately, a careless graduat

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Answer:

760 uM

Explanation:

<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>

The problem here is that the amount of magnesium ion remains the same irrespective of the volume.

Amount of magnesium in the growth medium = <em>molarity x volume</em>

= 97 x $10^{-6}$ x 47 x $10^{-3}$ = 4.559 x $10^{-6}$

Then, the volume reduced to 6.0 mL, the new molarity becomes;

<em>molarity = mole/volume </em>

= 4.559 x $10^{-6}$ /6 x $10^{-3}$ = 7.598333 x $10^{-4}$ M = 759.83333 uM

To the correct number of significant digits = 760 uM

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3 years ago

How many grams of He are necessary to fill a balloon having a volume of 4.50 × 10^{3} mL to a pressure of 1.14 × 10^{3} torr at

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The mass in grams of He are necessary to fill a balloon having a volume of 4.50×10³ mL to a pressure of 1.14×10³ torr at 25.0ºC is 1.104 grams.

<h3>How do we calculate the grams from moles?</h3>

Mass (W) in grams from moles (n) will be calculated by using the below equation:
n = W/M, where

M = molar mass

Moles of helium gas will be calculated by using the ideal gas equation:

PV = nRT, where

R = universal gas constant = 62.363 L.torr / K.mol

P = pressure of gas = 1.14×10³ torr

V = volume of gas = 4.50×10³ mL = 4.50 L

n = moles of gas = ?

T = temperature = 25.0ºC = 298 K

On putting these values on the ideal gas equation, we get

n = (1140)(4.5) / (62.363)(298) = 5130 / 18584

n = 0.276 moles

Now we convert this moles into mass by using the first equation as:

W = (0.276mol)(4g/mol) = 1.104 g

Hence required mass of helium gas is 1.104 grams.

To know more about ideal gas equation, visit the below link:

brainly.com/question/12873752

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2 years ago