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mash [69]
3 years ago
8

Identify the solute and solvent in each of the solutions. A. salt ( NaCl ) (NaCl) in water salt: water: B. air (a solution of 78

% N 2 , 78% N2, 21 % O 2 , 21% O2, and various other gases) N 2 : N2: O 2 : O2: C. a solution of 28 % 28% ethanol and 72 % 72% water ethanol: water: D. bronze (an alloy of 95 % 95% copper and 5 % 5% tin) copper:
Chemistry
1 answer:
stira [4]3 years ago
5 0

Answer:

A) solute - NaCl, solvent - water

B) solute - O2 and other gases, solvent - N2

C) solute - ethanol, solvent - water

D) solute - tin, solvent - copper

Explanation:

Solute(s) is/are the minor component(s) in a solution, dissolved in the solvent.

Solvent is the component of a solution that is present in the greatest amount. It is the substance in which the solute is dissolved.

If both solute and solvent exist in equal quantities (such as in a 50% ethanol, 50% water solution), the concepts of "solute" and "solvent" become less relevant, but the substance that is more often used as a solvent is normally designated as the solvent (in this example, water).

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According to Hund's rule of maximum spin multiplicity, how many singly-occupied orbitals are there in the valence shells of the
leva [86]

Answer:

A) carbon  - 2

B) cobalt  - 3

C) sulfur   - 2

D) fluorine   - 1

E) titanium   - 2

F) germanium  - 2

Explanation:

Hund's rule of maximum multiplicity:-

Firstly, every orbital which is present in the sublevel is singly occupied and then the orbital is doubly occupied.  

(A) Carbon.

The electronic configuration is -  

1s^22s^22p^2

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

(B) Cobalt.

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{7}4s^2

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 4 electrons will be paired in 2 orbitals and 3 orbitals will be singly filled in cobalt.

(C) Sulfur.

The electronic configuration is -  

1s^22s^22p^63s^23p^4

Thus, 3s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 2 electrons will be paired in 1 orbital and 2 orbitals will be singly filled in sulfur.

D) fluorine

The electronic configuration is -  

1s^22s^22p^5

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 4 electrons will be paired in 2 orbitals and 1 orbital will be singly filled in fluorine.

E) Titanium

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{2}4s^2

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 2 orbitals will be singly filled in titanium.

F) Germanium

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{10}4s^24p^2

Thus, 4s, 3d orbitals are fully filled and p orbital can singly filled 3 electrons. Thus, Germanium has 2 singly occupied orbitals.

4 0
3 years ago
Write the nuclear symbol of oxygen-16?​
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Answer:

Oxygen-16 (16o) is a stable isotope of oxygen, having 8 neutrons and 8 protons in its nucleus.

Explanation:

5 0
3 years ago
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Answer:

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Explanation:

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3 years ago
Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at
telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

H_2(g)+I_2(g)\rightarrow 2HI(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound            Pressure at equilibrium

H_2                                   61.8 atm

I_2                                    46.5 atm

HI                                  52.3 atm

Calculate the value of the equilibrium constant K_p for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant K_p for this reaction is, 0.952

Explanation :

The given chemical reaction :

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression of K_p for above reaction follows:

K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}

We are given:

P_{H_2}=61.8atm

P_{I_2}=46.5atm

P_{HI}=52.3atm

Putting values in above equation, we get:

K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952

Therefore, the value of equilibrium constant K_p for this reaction is, 0.952

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What is waste good for?
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