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3 years ago

You have 8 moles of a gas at 250 K in a 6 L container. What is the pressure of the gas? Show your work

Chemistry

1 answer:

Bogdan [553]3 years ago

7 0

Hello:

In this case, we will use the Clapeyron equation:

P = ?

n = 8 moles

T = 250 K

R = 0.082 atm.L/mol.K

V = 6 L

Therefore:

P * V = n * R * T

P * 6 = 8 * 0.082* 250

P* 6 = 164

P = 164 / 6

P = 27.33 atm

Hope that helps!

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what conclusions can be made about the relationship between metallic character and the atomic radius?

kolezko [41]

We have to get the relationship between metallic character and atomic radius.

Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.

If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.

With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.

With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.

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3 years ago

Which two body systems are most important in breaking down food and distributing nutrients throughout the body?

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Answer:

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2 years ago

HNO3 + H2O → H3O^+ + NO3<br> which ones are acids and which ones are bases

nlexa [21]

Answer:

HNO₃ and H₃O⁺ are acids

H₂O and NO₃⁻ are bases

Explanation:

The chemical equation is:

HNO₃ + H₂O → H₃O⁺ + NO₃⁻

There are several definitions of acid and bases: Arrhenius', Bronsted-Lowry's and Lewis'.

Bronsted-Lowry model defines and <em>acid</em> as a donor of protons, H⁺.

In the given equation HNO₃ is such substance: it releases an donates its hdyrogen to form the H₃O⁺ ion.

On the other hand, a <em>base</em> is a substance that accepts protons.

In the reaction shown, H₂O accepts the proton from HNO₃ to form H₃O⁺.

Thus, H₂O is a base.

In turn, on the reactant sides the substances can be classified as acids or bases.

H₃O⁺ contain an hydrogen that can be donated and form H₂O; thus, it is an acid (the conjugated acid), and NO₃⁻ can accept a proton to form HNO₃; thus it is a base (the conjugated base).

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3 years ago

When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+

8090 [49]

Answer:

D) 65.7%

Explanation:

Based on the reaction:

2H2(g)+O2(g)⟶2H2O(l)

<em>2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.</em>

<em />

To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.

<em>Theoretical yield:</em>

Moles of 5.58g H₂:

5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂

As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:

2.768 moles H₂O ₓ (18.015g / mol) =

49.86g H₂O is theoretical yield

<em>Percent yield:</em>

Percent yield = Actual yield / Theoretical yield ₓ 100

32.8g H₂O / 49.86g ₓ 100 =

65.7% is percent yield of the reaction

4 0

3 years ago

Calculate the pH after the addition of 10.0 mL of 0.240 M sodium hydroxide to 50.0 mL of 0.120 M acetic acid.

Goryan [66]

Answer:

pH = 4.58

Explanation:

The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>Moles that react:</em>

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles

CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

<em>Where [] is molarity of each species or moles</em>

<em />

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

<em />

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3 years ago