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Alexeev081 [22]
3 years ago
6

HELP ASAP!! how can nuclear energy be classified as an inexhaustible energy resource??

Chemistry
1 answer:
seropon [69]3 years ago
6 0

Answer:

No, and yes. As there is a finite amount of matter in the

universe, only so much can be converted to make energy. There will

be a limit, though it is a long, long way from where we are

Explanation:

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Can anybody please with this ohms law
mixer [17]

Answer:

feyyeyeuueuehehheueueieuhegebrbjrirororkoeorororproororororri9r

Explanation:

eejjeididodjdjrirr

7 0
3 years ago
What is the second quantum number of the 3p1 electron in aluminum 1s22s22p63s23p1?
Brut [27]
There are 4 quantum numbers that can be used to describe the space of highest probability an electron resides in.
First quantum number is the principal quantum number- n , states the energy level.
Second quantum number states the angular momentum quantum number - l,
states the subshell and the shape of the orbital
values of l for n energy shells are from 0 to n-1
third is magnetic quantum number - m, which tells the specific orbital.
fourth is spin quantum number - s - gives the spin of the electron in the orbital

here we are asked to find l for 3p1
n = 3
and values of l are 0,1 and 2
for p orbitals , l = 1
therefore second orbital for 3p¹ is 1.

5 0
3 years ago
Read 2 more answers
Balance the following reaction _ Al + _ Cl, → _ AICI,
Free_Kalibri [48]

Answer:

1 Al + 1 Cl -> 1 AlCl

Explanation:

It is already balanced.

3 0
3 years ago
What is different between margerine and butter in term of organic chemistry
Brut [27]

Answer:

The most important difference between the two is that butter is derived from dairy and is rich in saturated fats, whereas margarine is made from plant oils. ... If the margarine contains partially hydrogenated oils, it will contain trans fat, even if the label claims that it has 0 g.

Explanation:

(⌒_⌒;)

4 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
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