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ivanzaharov [21]
3 years ago
8

(Science)Pls help due by tmr night and I still have like 30 more questions ​

Chemistry
1 answer:
olga2289 [7]3 years ago
5 0

Answer:

A A B

Explanation:

I know that when it is slopped it is 90% always accelerating

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Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
galina1969 [7]

Answer:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Oxidized: Cd

Reduced: Ag

Explanation:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Cd → Cd²⁺  +  2e⁻      Half reaction oxidation

1e⁻ + Ag⁺ → Ag           Half reaction reduction

Ag changed oxidation number from +1 to 0

Cd changed oxidation number from 0 to +2

Let's ballance the electrons

( Cd → Cd²⁺  +  2e⁻ ) .1

( 1e⁻ + Ag⁺ → Ag ) .2

Cd + 2e⁻ + 2Ag⁺  → 2Ag +  Cd²⁺  +  2e⁻

Finally the ballance equation is:

Cd(s) + 2AgNO₃(aq)  → Cd(NO₃)₂ (aq) + 2Ag(s)

4 0
3 years ago
What does lithium 6 and lithium 7 look like
Marysya12 [62]
Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital. 
3 0
3 years ago
Hund's rule states that electrons must spread out within a given subshell before they can pair
Temka [501]

Answer:

Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.

Explanation:

If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.

6 0
2 years ago
Three isotopes of silicon have mass numbers of 28,29, and 30 with an average atomic mass of 28.086 what does this say about the
Sedbober [7]

Answer is: silicon isotope with mass number 28 has highest relative abundance, this isotope is the most common of these three isotopes.

Ar₁(Si) = 28; the average atomic mass of isotope ²⁸Si.  

Ar₂(Si) =29; the average atomic mass of isotope ²⁹Si.  

Ar₃(Si) =30; the average atomic mass of isotope ³⁰Si.  

Silicon (Si) is composed of three stable isotopes, ₂₈Si (92.23%), ₂₉Si (4.67%) and ₃₀Si (3.10%).

ω₁(Si) = 92.23%; mass percentage of isotope ²⁸Si.  

ω₂(Si) = 4.67%; mass percentage of isotope ²⁹Si.

ω₃(Si) = 3.10%; mass percentage of isotope ³⁰Si.

Ar(Si) = 28.086 amu; average atomic mass of silicon.  

Ar(Si) = Ar₁(Si) · ω₁(B) + Ar₂(Si) · ω₂(Si)  + Ar₃(Si) · ω₃(Si).  

28,086 = 28 · 0.9223 + 29 · 0.0467 + 30 · 0.031.

8 0
3 years ago
I need help with this
marshall27 [118]
What is it you need help on? there is nothing here????

ANSWER:
7 0
2 years ago
Read 2 more answers
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