All categories

4 years ago

.Photography first used at a crime scene. before the 1800s during the 1800s 1900-1988 1990-present

Chemistry

2 answers:

laila [671]4 years ago

6 0

Answer:

during the 1800s

Explanation:

Alphonse Bertillon (a French police officer) was the first to methodically photograph and document crime scenes. He is also the inventor of the mug shot.

Photographs at crime scenes and criminals started at mid-1800s and they were standardized at the end of that century when they came to be widely accepted as a forensic means of identification.

sveticcg [70]4 years ago

5 0

I'm going to say during the 1800s

You might be interested in

Cesium Bromide and Strontium Nitrate react to form Strontium Bromide and Cesium Nitrate.

natita [175]

2................ 2................ 2................

6 0

3 years ago

Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin

Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

CH₄ + 4Cl₂ --------→ CCl₄ + 4HCl

1 mol 4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of Cl₂ = 71 g/mol

mass of Cl₂

mass in grams = no. of moles x molar mass

mass of Cl₂ = 4 mol x 71 g/mol

mass of Cl₂ = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

mass in grams = no. of moles x molar mass

mass of CH₄= 1 mol x 16 g/mol

mass of CH₄ = 16 g

So,

284 g of Cl₂ combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂

Apply unity Formula

284 g of Cl₂ ≅ 16 g of methane ( CH₄ )

33.4 g of Cl₂ ≅ X g of methane ( CH₄ )

By cross multiplication

X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

methane (CH₄) = 1.882 g

5 0

3 years ago

The combination of sulfur dioxide, nitrogen oxide, and atmospheric moisture creates

sweet-ann [11.9K]

This combination creates acid rain

8 0

3 years ago

Read 2 more answers

HELP ME PLEASE!!!!

liq [111]

Answer:

1.D

2.A

3.C

4.A

Explanation:

6 0

3 years ago

Read 2 more answers

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

3 years ago