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schepotkina [342]
3 years ago
11

The volume of a gas is 17.5 mL at 29◦C and

Chemistry
1 answer:
mars1129 [50]3 years ago
8 0

Answer: V = 14.4 mL

Explanation:

In this case, we'll use the ideal gas formula:

PV = nRT

Where:

P: pressure in atm

V: volume in liter

n: moles

R: gas constant (In this case 0.082 L atm/mol K)

T: temperature in K

We have data for the gas at first, so we can actually calculate the moles used, which will be constant even at different temperature and pressure. Let's calculate the moles:

n = PV/RT

Before we do this, let's convert the volume and temperature:

T = 29 + 273 = 302 K

V = 17.5 / 1000 = 0.0175 L

Now the moles:

n = 0.863 * 0.0175 / 0.082 * 302 = 0.00061 moles

with these moles, let's convert the temperature to K:

T2 = 13 + 273 = 286 K

and to calculate the volume:

V = nRT/P

so:

V = 0.00061 * 0.082 * 286 / 0.994

V = 0.0144 L or simply 14.4 mL

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Which of the following statements is true?
Anna007 [38]

Answer:

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Explanation:

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So, this explains the answer.

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3 years ago
Why does a red object appear red?
NeTakaya

Answer:

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8 0
3 years ago
Read 2 more answers
A biochemist carefully measures the molarity of magnesium ion in of cell growth medium to be . Unfortunately, a careless graduat
denis23 [38]

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

6 0
3 years ago
Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base
lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 6.20*10^{-5}

So:

pKa = - log\:(Ka)

pKa = - log\:(6.20*10^{-5})

pKa = 5 - log\:6.20

pKa = 5 - 0.79

\boxed{pKa = 4.21}

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

pH = pKa + log\:\dfrac{[salt]}{[acid]}

pH = 4.21 + log\:\dfrac{0.130}{0.235}

pH = 4.21 + log\:0.55

pH = 4.21 + (-0.26)

pH = 4.21 - 0.26

\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

8 0
3 years ago
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