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3 years ago

Which of the following is an advance in technology that has helped us to increase environmental quality?

1 answer:

meriva3 years ago

3 0

OK, its deff not the last one b/c all that trash just goes into the air, overwater farm land would kill crops (no help there), Ive never heard of heating water to kill excess bacteria. There is ,however, a new technique called Bioremediation which is the use of living organism to clean up a medium i.e. air,soil, water. It would seem the 2nd answer is correct.

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Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0

3 years ago

What types of bonds or interactions are overcome when a nonmetal extended network melts?

STatiana [176]

Covalent bonds or interactions are overcome when a nonmetal extended network melts.

Typically, nonmetals form covalent bonds with one another. A polyatomic ion's atoms are joined by a form of link called covalent bonding. A covalent bond requires two electrons, one from each of the two atoms that are connecting.

One technique to depict the formation of covalent connections between atoms is with Lewis dot formations. The number of unpaired electrons and the number of bonds that can be formed by each element are typically identical. Each element needs to share an unpaired electron in order to establish a covalent bond.

Therefore, covalent bonds or interactions are overcome when a nonmetal extended network melts.

Learn more about covalent bonds here;

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4 0

2 years ago

4. In a solution of sugar water, sugar is the ______. (SPS6a; DOK2)

Rzqust [24]

Answer:

B. solute

Explanation:

6 0

4 years ago

Read 2 more answers

there is a total of seventeen constitutional isomers for the molecular formula c5h13n. draw the skeletal formula of all three co

MAXImum [283]

Amines are derivatives of Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;

Primary Amines:
In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.

Secondary Amines:
In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.

Tertiary Amines:
In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.

Below are three isomers of tertiary amines with molecular formula C₅H₁₃N.

5 0

3 years ago

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio

8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

$V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL$

4 0

3 years ago