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aleksandrvk [35]

4 years ago

9

How do people use bromine?

1 answer:

EastWind [94]4 years ago

6 0

Bromine is used in many areas such as the following agricultural chemicals, dyestuffs, insecticides, pharmaceuticals and chemical intermediates. Some of the uses are being phased out for environmental reasons, but new uses can be found with Bromine can be used for furniture foam, plastic casings for electronics and textiles to make them less flammable

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If a fluorine atom and nitrogen atom combine to form a compound, what type of bond will they form

dmitriy555 [2]

Answer:

I guess covalent bond is formed

4 0

3 years ago

What accounts for the large increase in volume when sugar burns?

Dimas [21]

I don’t know type it up yourself

8 0

3 years ago

How many liters of salt solution will be needed to provide 15 grams of salt if the concentration of the solution is 50 g/l?

asambeis [7]

Answer:

.30 l

Explanation:

15g/50g/l= .3l

6 0

3 years ago

Transuranium elements are artificial and may be prepared by a fusion process in which heavy nuclei are bombarded with light ones

Scilla [17]

Answer:

$_{97}^{345}\textrm{Bk}$

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

Am has an atomic number of 95. So correct reaction is:-

$^{343}_{95}\textrm{Am}+^{4}_{2}\textrm{He}\rightarrow ^A_Z\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

343 + 4 = A + 2

A = 345

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

95 + 2 = Z + 0

Z = 97

Hence, the isotopic symbol of unknown element is $_{97}^{345}\textrm{Bk}$

7 0

3 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,

1 ton=907.185 kg, thus,

$3.3480\times 10^{4} tons=3.037\times 10^{7} kg$

Thus, production of $CO_{2}$ will be $3.037\times 10^{7} kg/ day$ .

The specific volume of $CO_{2}$ is $0.0120 m^{3}/kg$ .

Volume of $CO_{2}$ produced per day can be calculated as:

$V=Specific volume\times mass$

Putting the values,

$V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day$

Thus, volume of $CO_{2}$ produced per year will be:

$V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year$

Thus, in 4 year volume of $CO_{2}$ produced will be:

$V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}$

8 0

3 years ago