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4 years ago

If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?

Chemistry

1 answer:

EastWind [94]4 years ago

4 0

Temperature and pressure are directly proportional. Pressure will also have to increase ( doubling).

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Determine the mass of an object that has a volume of 88.6 ml and a density of 9.77 g/ml.

Nostrana [21]

Given:
The volume is V = 88 mL
Density, d = 9.77 g/mL

Because mass = volume * density, therefore the mass of the object is
$m = (88.6 \, mL)*(9.77 \, \frac{g}{mL} )=865.62 \, g$

Answer: The mass is 865.2 g or 0.865 kg

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4 years ago

In the front of the room, there is a bottle that contains a 32.00 g sample of sulfur. This is

Vera_Pavlovna [14]

6.02x10^23 atoms is the answer

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3 years ago

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3 years ago

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Determine the nuclear composition (number of protons and neutrons) of the following isotopes.

miskamm [114]

Answer:

See below

Explanation: Vanadium 51 Protons 23 nutrons 28

Zinc 64 protons 30 nuetrons 34

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hope this helps god bless

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! =)

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3 years ago