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2 years ago

Which chemical helps reveal a damaged serial number of a gun

Chemistry

2 answers:

Hatshy [7]2 years ago

5 0

Answer: Fry's reagent

Used to see the small eteches in metal, carbon and a few other materials.

Lera25 [3.4K]2 years ago

4 0

Restoring a damaged serial number out from a gun is possible in forensic science. For example, a frame of a revolver containing an obliterated serial number, scientists can polish the metal using sand-paper type material to remove chips of metal from the surface. Then, chemical reagents or acids are applied to etch the metal until the number is restored. Chemicals used are based on the variety of metal surface, few examples include aluminum, zinc and steel alloys.

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A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes

insens350 [35]

0.208 is the specific heat capacity of the metal.

Explanation:

Given:

mass (m) = 63.5 grams 0R 0.0635 kg

Heat absorbed (q) = 355 Joules

Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K

cp (specific heat capacity) = ?

the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = $\frac{q}{mΔ T}$

c = $\frac{355}{63.5X 268.59}$

= 0.208 J/gm K

specific heat capacity of 0.208 J/gm K

The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.

5 0

2 years ago

If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the

jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce $\frac{68}{34}$ g of (NH₄)₂S

25g of NH₃ will produce $\frac{65}{34} X 25 g$ of (NH₄)₂S

= 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

4 0

2 years ago

The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG

Dmitrij [34]

Answer:

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP -----------> ADP + creatine phosphate

ΔG∘ = -12.6 KJ/mole = -12600 J/mole

ΔG = -0.1 KJ/mole = -100 J/mole

[Creatine phosphate] = 25 mM = 25*10^-3 M

[Creatine] = 17 mM = 17*10^-3 M

[ATP] =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600 = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500 = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc = 6.44*10^-3

But,

Qc = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3 = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

= 2.1896 *10^-5 M

= 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

3 0

2 years ago

Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r

Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

$24 \times 3600 sec$

Hence, total charge passed daily is calculated as follows.

$150,000 \times 24 \times 3600 sec$

And, number of Faraday of charge is as follows.

$\frac{150,000 \times 24 \times 3600 sec}{96500}$

= 134300.52 F

The oxidation state of aluminium in $Al_{2}O_{3}$ is +3.

$Al^{3+} + 3e^{-} \rightarrow Al(s)$

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

$\frac{134300.52 F}{3} \times \frac{89}{100}$

= 39842.487 mol

or, = $3.9842 \times 10^{4} mol$

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

Mass in grams = $3.9842 \times 10^{4} mol \times 27$

= $107.57 \times 10^{4} g$

= 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

8 0

2 years ago

Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod

son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0

2 years ago