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Diano4ka-milaya [45]
3 years ago
13

How can endangered species be saved ?

Chemistry
2 answers:
labwork [276]3 years ago
5 0

Answer:

There are lots of methods.

Explanation:

Usually, animals like pandas live a shorter lifespan in the wild than in captivity. A little fact, there is only one brown panda in the entire world, so it would be very, very rare to see one. The Smithsonian National Zoo, for example, are working to protect pandas, as well as other species.

MatroZZZ [7]3 years ago
3 0

Answer:

Protection

Explanation:

Different types of supporting by wildlife habitats, Joining of Endangered species programs.

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Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all
Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

  • n = 4.
  • l  = 1.
  • m_l = 0.
  • Either m_s = 1/2.

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the 4s orbital (fourth s orbital.)

The quantum number n (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, n = 4 for this electron.

The quantum number l (the angular momentum quantum number) specifies the shape (s, p, d, etc.) of an electron. l = 1 for s\! orbitals (such as the one that contains this electron.

Quantum numbers n and l specify the shape of an orbital. On the other hand, the magnetic quantum number m_l specifies the orientation of these orbitals in space.

However, s orbitals are spherical. Regardless of the value of n, the only possible m_l value for electrons in s\! orbitals is m_l = 0.

The spin quantum number m_s distinguishes between the two electrons in an orbital. The two possible values of m_s \! are (+1/2) and (-1/2). Typically, the first electron in an orbital is assigned an upward (\uparrow) spin, which corresponds to m_s = (+1/2).

5 0
3 years ago
A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached Ne
bulgar [2K]

Answer:

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

Explanation:

The acidity of a solution depends on the concentration of H⁺ ions ([H⁺]). We can calculate this concentration from the pH using the following expression.

pH = -log ([H⁺])

American Midwest

pH = -log ([H⁺])

5.02 = -log ([H⁺])

[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M

New England

pH = -log ([H⁺])

4.66 = -log ([H⁺])

[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M

The ratio of concentrations is:

\frac{2.19 \times 10^{-5} M  }{9.55 \times 10^{-6} M} =2.29

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

4 0
3 years ago
HELP ME QUICK
Sophie [7]
Quick answer: Exothermic reactions mean that heat has escaped the system. The answer is A. Heat is released.
3 0
3 years ago
Read 2 more answers
What is the percent cr, by mass, in the steel sample? express your answer numerically as a percentage?
djverab [1.8K]
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−

Assuming all the Cr is contained in the BaCrO4 at the end. 
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
8 0
3 years ago
Read 2 more answers
Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their
dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Pure Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Br and Br,

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

                                                   ________

                    E.N Difference             0.00         (Non Polar/Pure Covalent)

 

For N and O,

                    E.N of Oxygen      =   3.44

                    E.N of Nitrogen     =   3.04

                                                   ________

                    E.N Difference             0.40           (Non Polar/Pure Covalent)

 

For P and H,

                    E.N of Hydrogen       =   2.20

                    E.N of Phosphorous  =   2.19

                                                              ________

                    E.N Difference                  0.01          (Non Polar/Pure Covalent)

 

For K and O,

                    E.N of Oxygen          =   3.44

                    E.N of Potassium      =   0.82

                                                   ________

                    E.N Difference                2.62              (Ionic)

6 0
4 years ago
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