The mass of NaCl needed for the reaction is 91.61 g
We'll begin by calculating the number of mole of F₂ that reacted.
- Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1.5 × 12 = n × 0.0821 × 280
18 = n × 22.988
Divide both side by 22.988
n = 18 / 22.988
n = 0.783 mole
Next, we shall determine the mole of NaCl needed for the reaction.
F₂ + 2NaCl —> Cl₂ + 2NaF
From the balanced equation above,
1 mole of F₂ reacted with 2 moles of NaCl.
Therefore,
0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.
Finally, we shall determine the mass of 1.566 moles of NaCl.
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = mole × molar mass
Mass of NaCl = 1.566 × 58.5
Mass of NaCl = 91.61 g
Therefore, the mass of NaCl needed for the reaction is 91.61 g
Learn more about stiochoimetry: brainly.com/question/25830314
The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.
An ion can be cation (positively charged) or anion (negatively charged).
Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).
The remaining elements are completed as follows to attain octet structure;
<u>Element</u>--<u>valence electron</u>--<u>electrons to gain</u>--<u>electrons to lose</u>--<u>ion formed</u>
O ------------ 6 ---------------------- 2 ------------------------ none -------------- 
Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ 
Br ----------- 7 --------------------- 1 ------------------------ none --------------- 
S ------------ 6 ----------------------- 2 ------------------------ none --------------- 
Cl ------------ 7 ----------------------- 1 ------------------------ none ----------------
K -------------- 1 ----------------------- none ----------------------- 1 ------------------ 
Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- 
Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- 
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<h3>
Answer:</h3>
2000 atoms
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Explanation:</h3>
We are given the following;
Initial number of atoms of radium-226 as 8000 atoms
Time taken for the decay 3200 years
We are required to determine the number of atoms that will remain after 3200 years.
We need to know the half life of Radium
- Half life is the time taken by a radio active material to decay by half of its initial amount.
- Half life of Radium-226 is 1600 years
- Therefore, using the formula;
Remaining amount = Original amount × 0.5^n
where n is the number of half lives
n = 3200 years ÷ 1600 years
= 2
Therefore;
Remaining amount = 8000 atoms × 0.5^2
= 8000 × 0.25
= 2000 atoms
Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.