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3 years ago

How many moles are in 1.204 x 10^24 atoms of gold?

Chemistry

1 answer:

Arturiano [62]3 years ago

8 0

Answer:

<h3>The answer is 2 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 1.204 × 10²⁴ atoms of gold

We have

$n = \frac{1.204 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>2 moles</h3>

Hope this helps you

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the Temperature of a sample of an ideal gas in a sealed 5.0 l container is raised from 27 oC to 77oC. If the initial pressure of

matrenka [14]

Answer:
The pressure increases to 3.5 atm.

Solution:
According to Gay-Lussac's Law, " At constant volume and mass the pressure of gas is directly proportional to the applied temperature".

For initial and final states of a gas the equation is,

P₁ / T₁ = P₂ / T₂
Solving for P₂,
P₂ = P₁ T₂ / T₁ ----- (1)

Data Given;
P₁ = 3 atm

T₁ = 27 °C + 273 = 300 K

T₂ = 77 °C + 273 = 350 K

Putting values in eq. 1,

P₂ = (3 atm × 350 K) ÷ 300 K

P₂ = 3.5 atm

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3 years ago

A buffer is prepared by adding 12.0g of ammonium chloride (NH4Cl) to 250mL of 1.00 M NH3 solution?The pH is 9.3 Write the net io

solniwko [45]

Answer:- $NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)$

Explanations:- The solution we have is a buffer solution and we know that a buffer solution resists a change in its pH if a strong acid or base is added to it.

Here, the buffer solution we have is of a weak base and it's conjugate acid. So, a strong acid(nitric acid) is added to this buffer then it reacts with the base present in the buffer so that the acid could be neutralized. This is called buffer action.

The net ionic equation is written as:

$NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)$

Note that $HNO_3$ is a strong acid and nitrate ion is the spectator ion so it is not included in the net ionic equation.

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2 years ago

If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?

olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

HA H₃O⁺ A⁻

Initial, M 0.40 0 0

Change , M -x +x +x

Equilibrium, M 0.40 - x x x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8