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uranmaximum [27]
4 years ago
10

.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
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What type of reaction is Pb(NO3)2 (aq) + 2KI (aq) →PbI2 + 2KNO3 (aq)
kifflom [539]

Answer:

Precipitation Reactions

They contain two aqueous reactants, one aqueous product, and one solid product. In this reaction, two soluble products, Pb(NO3)2 and KI, combine to form one soluble product, KNO3, and one insoluble product, PbI2. This is a precipitation reaction, and PbI2 is the precipitate.

4 0
3 years ago
What volume of 0.08892 M HNO3 is required to react completetly with 0.2352 g of potassium hydrogen phosphate?
galina1969 [7]

Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

Molarity of the HNO₃ Solution = 0.08892 M

Now,

From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃

The number of moles of 0.2352 g of potassium hydrogen phosphate

= Mass / Molar mass

also,

Molar mass of potassium hydrogen phosphate

= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol

Number of moles = 0.2352 / 174.15 = 0.00135 moles

thus,

The number of moles of HNO₃ required for  0.00135 moles

= 2 ×  0.00135 mol of HNO₃

= 0.0027 mol of HNO₃

Now,

Molarity = Number of Moles / Volume

thus,

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0.08892 = 0.0027 / Volume

or

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8 0
3 years ago
Read 2 more answers
What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
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Mass of carbon dioxide produced  
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Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
4 0
3 years ago
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