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Flauer [41]
3 years ago
8

For the IMAGINARY molecular compound fluorine tri nitride having the formula: FN3

Chemistry
2 answers:
Alexxx [7]3 years ago
5 0

Answer:

The answer to your question is the letter D)

Explanation:

Data

       FN₃

Process

-When we are forming molecules, the oxidation numbers of the atoms cross.

- This means that if the oxidation number of fluorine is ⁺³, the oxidation number of Nitrogen is ⁻¹, and when we cross the oxidation numbers, the 3 will be the number of molecules of nitrogen and the number 1 will be the number of molecules of fluorine.

Conclusion

There is 1 atom of fluorine for every 3 atoms of nitrogen.                              

Sergio [31]3 years ago
5 0
The answer is D, There are 1 atoms of fluorine for every 3 atoms of nitrogen
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antiseptic1488 [7]
I feel u r working with the chapter Mechanism of liquid
but I could understand ur question....
3 0
4 years ago
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If I were to draw an electron configuration for a bromine (Br) atom, what is the highest energy level I would reach?
dimaraw [331]

Answer:

4p

Explanation:

3s < 3p < 4s < 4p

This arrangement is in order of increasing energy.

This explains that, 4p is the highest energy level you can reach.

3 0
3 years ago
Cream of tartar creates a white-purple flame when burned. What can you hypothesize is a component of the cream
antiseptic1488 [7]

Potassium can be a component of the cream.

<h3>Why does potassium show white-purple flame when burned?</h3>

An element's presence can be determined using a flame test. The potassium in cream of tartar is what gives it its white-purple color.

Low ionization enthalpy elements allow the flame to display color. Due to low ionization enthalpy, when an element is heated, its valence electrons are quickly excited to higher orbits and emit light as they return to their original orbit. Salts of potassium emit a white-purple color when ignited. The cream of tartar's chemical formula is KC₄H₅O₆.

Learn more about flame test here:

brainly.com/question/22043426

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4 0
3 years ago
A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden
Luda [366]

Answer:

Concentration of Cr_2O_7^{2-} = 0.03101 M

Concentration of MnO_4^- = 0.03721 M

Explanation:

A)

The reduction for Cr_2O_7^{2-} is;

Cr_2O_7^{2-} + 14 H ^+ _{(aq)}  + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}

Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-

6 moles of Cu ^+ = 1 mole of Cr_2O_7^{2-}

number of moles of Cu reacted = \frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }

number of moles of Cu reacted = \frac{0.4749}{63.55}

number of moles of Cu reacted = 0.00747 mole

number of moles of Cr_2O_7^{2-}reacted = \frac{0.00747}{6}

number of moles of Cr_2O_7^{2-}reacted = 0.001245 mole

Concentration of Cr_2O_7^{2-} = \frac{number \ of moles }{Volume}

Given that the volume = 40.15 mL = 40.15 *10^{-3}; we have:

Concentration of Cr_2O_7^{2-} = \frac{0.001245}{40.15*10^{-3}}

Concentration of Cr_2O_7^{2-} = 0.03101 M

B)

The reduction for MnO_4^- is;

MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O

Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-

5 moles of Cu ^+ = 1 mole of Cr_2O_7^{2-}

number of moles of Cu reacted = \frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }

number of moles of Cu reacted = \frac{0.4749}{63.55}

number of moles of Cu reacted = 0.00747 mole

number of moles of MnO_4^- reacted = \frac{0.00747}{5}

number of moles of MnO_4^- reacted = 0.001494 mole

Concentration of MnO_4^- = \frac{number \ of moles }{Volume}

Given that the volume = 40.15 mL = 40.15 *10^{-3}; we have:

Concentration of MnO_4^- = \frac{0.001494 }{40.15*10^{-3}}

Concentration of MnO_4^- = 0.03721 M

3 0
4 years ago
1. Rubidium is a soft, silvery-white metal that has two common isotopes, 35Rb and Rb. The atomic mass of
xxTIMURxx [149]

Answer:

85.076572

Explanation:

= (84.911789×0.7217) + (86.909183×0.2738)

= 85.076572

8 0
3 years ago
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