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Nana76 [90]
3 years ago
9

The atomic number of lead (Pb) is 82. Its mass number is 207. How many protons, neutrons, and electrons does this neutral atom o

f lead (Pb) contain?
Chemistry
1 answer:
AveGali [126]3 years ago
7 0
An element’s atomic number is equal to the number of protons in that element’s nucleus. The mass number is the total number of an atom’s protons and neutrons. Protons have a positive charge; electrons have a negative charge; and neutrons are electrically neutral.

Putting it all together, given that the atomic number of lead is 82, the number of protons a lead atom contains is 82. The number of neutrons would be the difference between 207 and 82, or 125 neutrons. Finally, since you have a neutral atom, there must be an equal number of electrons as the number of protons—that is, 82 electrons.

Thus, you’ve got 82 protons, 125 neutrons, and 82 electrons.
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Answer:

I think the answer would be b, sorry if I'm wrong(EDIT: ITS ACTUALLY AAAAA)

5 0
2 years ago
The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is
Feliz [49]
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
5 0
3 years ago
2. if 0.20 m fe3 had been used instead of 0.020 m fe3 , how would the numerical value of the rate constant and the activation en
dezoksy [38]

the calculated value is Ea is 18.2 KJ and A is 12.27.

According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.

At 500K, K=0.02s−1

At 700K, k=0.07s −1

The Arrhenius equation can be used to calculate Ea and A.

RT=k=Ae Ea

lnk=lnA+(RT−Ea)

At 500 K,

ln0.02=lnA+500R−Ea

500R Ea (1) At 700K lnA=ln (0.02) + 500R

lnA = ln (0.07) + 700REa (2)

Adding (1) to (2)

700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.

=ln [0.02/0 .07]

Ea= 2/35×100×8.314×1.2528

Ea =18227.6J

Ea =18.2KJ

Changing the value of E an in (1),

lnA=0.02) + 500×8.314/18227.6

= (−3.9120) +4.3848

lnA=0.4728

logA=1.0889

A=antilog (1.0889)

A=12.27

Consequently, Ea is 18.2 KJ and A is 12.27.

Learn more about Arrhenius equation here-

brainly.com/question/12907018

#SPJ4

5 0
1 year ago
Which of the subshells in the electron configuration of hf behave as core orbitals?
Nesterboy [21]
<span>Of all the sub-shells shown ( 1s ,2s ,2p ,3s ,3p ,4s ,3d ,4p ,5s ,4d ,5p ,6s ,4f ,5d) the ones that act as core orbital of HF (Hydrogen Fluoride) is 6s and 5d</span>
3 0
3 years ago
A is a d-aldohexose and B is an L-aldohexose. they afford the same optically active aldaric acid after dilute nitric acid oxidat
Alexeev081 [22]

Answer:

D-Glucose and L-Glucose

Explanation:

Aldohexose are the sugars which have six number of carbons and ends up in having an aldehyde group at one end. When dilute nitric acid is treated with any of them, the molecule gets oxidized (gets oxygen) and therefore turns into carboxylic acid.

The name of A is D-Glucose, and B is L-Glucose. Please find the structural formula attached.

8 0
3 years ago
Read 2 more answers
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