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sleet_krkn [62]
3 years ago
9

Please I need help! It is timed!

Chemistry
2 answers:
frez [133]3 years ago
5 0

Answer:

From point D to G, the potential energy increases

Explanation:

CaHeK987 [17]3 years ago
3 0

Answer: I think your answer is correct

Explanation:

I haven't studied this yet but from what I see it's correct because from G to D it decreases

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While you are doing your homework, a friend looks over your shoulder and sees electron configurations on your paper. "Whats That
zzz [600]

Answer:

The energy of atomic orbitals increases as the principal quantum number, n, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of l differ so that the energy of the orbitals increases within a shell in the order s < p < d < f.

Explanation:

7 0
3 years ago
HELPP nowww plssss!!!!
yKpoI14uk [10]
The answer is #2: gas
4 0
3 years ago
Can anyone help me with this asap?
tresset_1 [31]

Answer:

Explanation:

1) During the diagnosis of thyroid disease a 10 g sample of I-131 is used. After a period of 32 days, how much sample is still radio active.?

Answer:

0.625 g

Explanation:

HL = Elapsed time/half life

32 days/8 days = 4

At time zero = 10 g

At 1st half life = 10/2 = 5 g

At 2nd half life = 5/2 = 2.5 g

At 3rd half life = 2.5 /2 = 1.25 g

At 4th half life = 1.25 / 2 = 0.625 g

After 32 days still 0.625 g of I-131 remain radioactive.

2) what was the original mass of sample Tc-99 that was used to locate the brain tumor If 0.10 g of a sample remains after 30 days? (half life 6 days)

Answer:

0.32 g.

Explanation:

Half life = time elapsed / HL

Half life = 30 days / 6 days = 5

At 5th half life = 0.10 g

At 4th half life = 0.2 g

At 3rd half life =  0.4 g

At 2nd  half life = 0.8 g

At 1st half life = 0.16 g

At time zero = 0.32 g

The original amount was 0.32 g.

3) write the beta decay equation of I-131?

Equation:

¹³¹I₅₃  →  ¹³¹Xe₅₄ + ⁰₋₁e

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiation

¹³¹I₅₃  →  ¹³¹Xe₅₄ + ⁰₋₁e

The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.

3 0
3 years ago
A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.
slavikrds [6]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant ({ \alpha}) is :

\qquad \sf  \dashrightarrow \:  \alpha =  \frac{dissociation \:  \: percentage}{100}

\qquad \sf  \dashrightarrow \:  \alpha =  \frac{9.4}{100}  = 0.094

Now, at initial stage :

  • \textsf{ Conc of HA = 0.200 M}

  • \textsf{Conc of H+ = 0 M}

  • \textsf{Conc of A - = 0 M}

At equilibrium :

  • \textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}

  • \textsf{Conc of H+ = 0.094(0.200)  = 0.0188 M}

  • \textsf{Conc of A - = 0.094(0.200)  = 0.0188 M}

Now, we know :

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}

( big brackets represents concentration )

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}}

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}}

\qquad \sf  \dashrightarrow \: { K_a \approx 0.00195 }

\qquad \sf  \dashrightarrow \:  {K_a \approx 1.9 × {10}^{-3} }

7 0
2 years ago
Important question, please answer(worth 64 points.)
Blababa [14]

Answer:

Is is talking about h20 water?

Explanation:

4 0
3 years ago
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