Answer:
OPTION (A) : Testing a rock sample for gold content
Explanation:
For testing a rock sample of gold content you will need a Chemist. To test the material, the sample is rubbed on black stone which will leave a mark on the stone. This mark is tested by applying aqua fortis i.e nitric acid on the mark. If the mark gets dissolve then the material is not gold. If the mark sustain the it is further tested by applying aqua regia i.e nitric acid and hydrochloric acid which will prove the sample is of gold if it gets dissolve on using hydrochloric acid. The purity of the sample can be checked by differing the concentration of the aqua regia and comparing it with the gold material of the known purity.
Answer is: acid-base indicator or pH indicators.
Acid-base indicators are usually weak acids or bases and they are chemical<span> detectors for hydrogen or hydronium cations.</span>
Example for acid-base indicator is phenolphthalein (molecular formula C₂₀H₁₄O₄). Phenolphthalein is <span>colorless in </span>acidic<span> solutions and pink in </span>basic<span> solutions.
Another example is m</span><span>ethyl orange. It is red colour in acidic solutions and yellow colour in basic solutions.</span>
If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
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Answer:
Volume will goes to increase.
Explanation:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.
Now we will put the suppose values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 4.5 L × 348 K / 298 k
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Hence prove that volume increase by increasing the temperature.
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
