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2 years ago

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

Chemistry

1 answer:

motikmotik2 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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Explanation:

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1 year ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

2 years ago

Who discovered significance of 6.022 x 10 to the 23 power

Masja [62]

6.02 x 10^23 is Alvogrado's number (Amedio Alvogrado is the name). This formula is determined to find the amount of particles within a substance by a mole.
I hope this helps!

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2 years ago