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3 years ago

You weigh a 24K gold chain and find that it weighs 400 grams. Determine home many moles of gold atoms you have

Chemistry

1 answer:

makkiz [27]3 years ago

4 0

Answer:

2.03 moles of Gold

Explanation:

Gold is one of the most precious metal metal used in many applications and mainly as a jewellery. In terms of purity it is categorized in Karats. 24 Karat is considered the purest Gold (i.e. 100 % Gold) while other Karats (14, 18, 22 e.t.c) are alloys with other metals and gyms.

Data Given:

Mass of Gold = 400 g

A.Mass of Gold = 196.97 g.mol⁻¹

Calculate Moles of Gold as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 400 g ÷ 196.97 g.mol⁻¹

Moles = 2.03 moles of Gold

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.Find the pH if the [H3O+1] is 6.2  10-10 M

pishuonlain [190]

Answer:

The pH is 9,21.

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and higher than 7 to 14 basic. It is calculated as

pH = -log (H 30+)

pH= -log (6,2 x 10^-10)

3 0

3 years ago

Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b

Alinara [238K]

Answer:

Explanation:

The cell reaction properly written is shown below:

Cu|Cu²⁺ $_{aq}$ || Ag⁺ $_{aq}$ | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

Oxidation half:

Cu $_{s}$ ⇄ Cu²⁺ $_{aq}$ + 2e⁻

At the anode, oxidation occurs.

Reduction half:

Ag⁺ $_{aq}$ + 2e⁻ ⇄ Ag $_{s}$

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

Cu $_{s}$ ⇄ Cu²⁺ $_{aq}$ + 2e⁻

Ag⁺ $_{aq}$ + e⁻ ⇄ Ag $_{s}$

we multiply the second reaction by 2 to balance up:

2Ag⁺ $_{aq}$ + 2e⁻ ⇄ 2Ag $_{s}$

The net reaction equation:

Cu $_{s}$ + 2Ag⁺ $_{aq}$ + 2e⁻⇄ Cu²⁺ $_{aq}$ + 2e⁻ + 2Ag $_{s}$

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

Cu $_{s}$ + 2Ag⁺ $_{aq}$ ⇄ Cu²⁺ $_{aq}$ + 2Ag $_{s}$

6 0

3 years ago

1. a.) Calculate the wavelength of light which has a frequency of 5.25 x 10 14 Hz.

Scilla [17]

<h3>Answer:</h3>

5.71 × 10² nm

<h3>Explanation:</h3>

The product of wavelength and frequency of a wave gives the speed of the wave.

Therefore;

Velocity of wave = Wavelength × Frequency

c = f ×λ

In our case;

Frequency = 5.25 × 10^14 Hz

Speed of light = 2.998 × 10^8m/s

But;

λ = c ÷ f

= 2.998 × 10^8m/s ÷ 5.25 × 10^14 Hz

= 5.71 × 10^-7 m

But; 1 M = 10^9 nm

Therefore;

wavelength = 5.71 × 10^-7 × 10^9

= 5.71 × 10² nm

The wavelength of light wave 5.71 × 10² nm

3 0

3 years ago

Which of the following is an example of mechanical weathering

777dan777 [17]

The answer is A-Gravity. Why? This is because B and C are examples of chemical weathering and not mechanical. Although choice D may seem viable lava intrusion is not a direct cause of mechanical weathering although lava pushing upward may help in aiding mechanical weathering it would not be considered a big enough cause, thus gravity is the correct answer.

8 0

3 years ago

What is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest whole-number coeffici

rodikova [14]

Answer:

Option (B) 7

Explanation:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

To know the coefficient of O2 in the above equation, let us balance the equation.

The above equation can be balance as follow:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 3 in front of CO2 as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + H2O(l)

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + 3H2O(l)

There are a total of 4 atoms of O on the left side and a total of 9 atoms on the right side. It can be balance by putting 7/2 in front of O2 as show below:

C3H6O2(l) + 7/2O2(g) → 3CO2(g) + 3H2O(l)

Multiply through by 2

2C3H6O2(l) + 7O2(g) → 6CO2(g) + 6H2O(l)

Now, the equation is balanced.

From the balanced equation above, the coefficient of O2 is 7.

7 0

3 years ago