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maxonik [38]
3 years ago
9

Pick a one world factor to describe the benefits and limitations of Bromine

Chemistry
1 answer:
motikmotik3 years ago
7 0

A one world factor to describe the benefits and limitations of Bromine are water purification compound.

Explanation:

Benefits of bromine

  • Bromine is used to make organobromo compounds.
  • Dibromoethane is an agent for leaded gasoline, largely stepped out due to environmental considerations.  
  • In fire extinguishers, pharmaceutical products and insecticides, organobromines are used.  
  • Bromine applied in dyes production, water purification compounds, fumigants, sanitizes, medicinal,  photography, flame proofing agents.
  • In citrus beverages, bromine used as emulsifier.

Limitations of bromine

  • Organic bromines are used as protecting and disinfecting agents, because microorganisms has its damaging effect. Once it is used in greenhouses and on farmland they can easily wash off to surface water, which has negative health effects on algae, daphnia, lobsters and fishes.
  • The animals effected on DNA damage and nerve damage which can leads to development of cancer.
  • Organic bromines can damage organs such as kidneys, liver, lungs.
  • Ethylene bromine can cause cancer.

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L
slega [8]

Answer:

5.41 g

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

Moles=0.033 \times {0.45}\ moles

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.01485\ g= \frac{Mass}{342.39\ g/mol}

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

<u>Answer -  5.41 g</u>

5 0
3 years ago
Why is water able to stick to the side of the glass?
Ira Lisetskai [31]
The answer is surface tension 
8 0
3 years ago
Read 2 more answers
A reasonable pka for a weak base is:<br><br> 10.3<br><br> 0.9<br><br> 5.8<br><br> 7.4<br><br> 13.1
andriy [413]

Answer:

0.9

Explanation:

The pka represents the force by which the molecules need to dissociate for the acids ,

Hence , lower the pka stronger will be the acid and that therefore will  dissociate completely and vice versa , for a weak acid higher the pka .

And in case of a base , it will be completely reversed , lower pKa , weaker base ,

and higher pKa , stronger base .

From the data of the question ,

0.9 is the lowest value of the pKa , hence , weakest base .

7 0
3 years ago
True or False
DedPeter [7]

Answer:

true

Explanation:

4 0
3 years ago
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