We read your photo as:

The following refers to element Y. The table below shows the relative atomic masses and percentage abundance of isotopes. (a) Element V has two isotopes. Two thirds of V and one third of V and one third of atomic mass of element V? What is the

Chemistry

ScanSolve AI Answer

Answers

4 parts
2

2.8.1\text{2.8.1}

2.1\text{2.1}

3

63.54\text{63.54}

4

55\text{55}

26\text{26}

5

26\text{26}

69.09%\text{69.09\%}

30.91%\text{30.91\%}

Step-by-step solution

9 steps

Haya lasty, tuendelee na maswali mengine.

Part 2· 3 steps

  • Element A: Masi ya atomiki = 23, Neutroni = 12.
  • Element B: Masi ya atomiki = 7, Neutroni = 4.

  1. Kokotoa idadi ya protoni kwa kila elementi.

    • Idadi ya protoni = Masi ya atomiki - Idadi ya neutroni.
    • Kwa Element A: Idadi ya protoni = 23−12=1123 - 12 = 11.
    • Kwa Element B: Idadi ya protoni = 7−4=37 - 4 = 3.

  2. Andika mpangilio wa elektroni.

    • a) Kwa Element A (protoni 11): Mpangilio wa elektroni ni 2.8.1\boxed{\text{2.8.1}}.
    • a) Kwa Element B (protoni 3): Mpangilio wa elektroni ni 2.1\boxed{\text{2.1}}.

  3. Tambua elementi yenye nishati ya ioni (ionization energy) kubwa zaidi na kueleza.

    • b) Elementi B ina nishati ya ioni kubwa zaidi.
    • Maelezo: Elementi B ina maganda machache ya elektroni (maganda 2) ikilinganishwa na Elementi A (maganda 3). Elektroni ya nje kabisa katika Elementi B iko karibu zaidi na kiini na inakabiliwa na mvuto mkali zaidi wa kiini, hivyo inahitaji nishati zaidi kuiondoa.

Part 3

63.54\text{63.54}

  • M1: Masi ya atomiki = 62.93, Asilimia ya wingi = 69.09%
  • M2: Masi ya atomiki = 64.93, Asilimia ya wingi = 30.91%

  1. Tumia fomula ya kukokotoa masi ya atomiki ya wastani.

    Masi ya Atomiki ya Wastani=(Masi ya Isotope 1×%Wingi 1)+(Masi ya Isotope 2×%Wingi 2)100\text{Masi ya Atomiki ya Wastani} = \frac{(\text{Masi ya Isotope 1} \times \% \text{Wingi 1}) + (\text{Masi ya Isotope 2} \times \% \text{Wingi 2})}{100}

  2. Badilisha thamani na kokotoa.

    Masi ya Atomiki ya Wastani=(62.93×69.09)+(64.93×30.91)100\text{Masi ya Atomiki ya Wastani} = \frac{(62.93 \times 69.09) + (64.93 \times 30.91)}{100} Masi ya Atomiki ya Wastani=4347.1037+2006.9113100\text{Masi ya Atomiki ya Wastani} = \frac{4347.1037 + 2006.9113}{100} Masi ya Atomiki ya Wastani=6354.015100\text{Masi ya Atomiki ya Wastani} = \frac{6354.015}{100} Masi ya Atomiki ya Wastani=63.54\text{Masi ya Atomiki ya Wastani} = \boxed{\text{63.54}}

Part 4

55\text{55}

  • Elementi V ina isotopi mbili.
  • Theluthi mbili ya V ina masi ya atomiki 54.
  • Theluthi moja ya V ina masi ya atomiki 57.

  1. Kokotoa masi ya atomiki ya wastani ya elementi V.

    • a) Masi ya Atomiki ya Wastani = (23×54)+(13×57)(\frac{2}{3} \times 54) + (\frac{1}{3} \times 57)
    • Masi ya Atomiki ya Wastani = 36+1936 + 19
    • Masi ya Atomiki ya Wastani = 55\boxed{\text{55}}

  2. Kokotoa idadi ya protoni katika isotopi B.

    • b) Kwa isotopi C: Masi ya isotopi = 57, Neutroni = 31.
    • Idadi ya protoni katika C = Masi ya isotopi - Neutroni = 57−31=2657 - 31 = 26.
    • Kwa kuwa A, B, na C ni isotopi za elementi Y, zote zina idadi sawa ya protoni.
    • Kwa hivyo, idadi ya protoni katika isotopi B ni 26\boxed{\text{26}}.

Part 5

26\text{26}

  • Kutokana na swali la 4(b), isotopi C ina neutroni 31 na masi ya isotopi 57.
  • Jedwali linaonyesha masi za atomiki na asilimia ya wingi wa isotopi L1 na L2 za elementi L.

  1. Pata idadi ya protoni katika isotopi B.

    • Kama ilivyokokotolewa katika swali la 4(b), idadi ya protoni katika isotopi C ni 57−31=2657 - 31 = 26.
    • Kwa kuwa isotopi B ni ya elementi Y, ina idadi sawa ya protoni.
    • Idadi ya protoni katika isotopi B ni 26\boxed{\text{26}}.

  2. Andika asilimia ya wingi wa isotopi L1 na L2.

    • Kutoka jedwali lililotolewa:
      • Asilimia ya wingi wa isotopi L1 ni \boxed{\text{69.09%}}.
      • Asilimia ya wingi wa isotopi L2 ni \boxed{\text{30.91%}}.
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Chemistry Questions

QA) Define: i) Atomic number ii) Mass number b) Calculate the relative atomic mass of element X given…QAn element X has two naturally occurring isotopes of 75% and 25% and 37x with a relative abundance…QCalculate the relative atomic mass of element X, given the isotopes' mass numbers and abundances.QThe table below shows the number of protons, neutrons, electrons and the mass numbers of some atoms…QGiven an element with an average atomic mass of 35.5 and isotopes of mass numbers 35 and 37…
See all Chemistry questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp