Answer:
126.8, Iodine
Explanation:
- mass ×abundance/100
- (126.9045×80.45/100)+(126.0015×17.23/100)+(128.2230×2.23/100)
- 102.1+21.7+3=126.8
<em>IODINE</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomic</em><em> </em><em>mass</em><em> </em><em>of</em><em> </em>126.8 or 126.9
<h3>
Answer:</h3>
0.35 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial volume as 35.0 mL or 0.035 L
- Initial molarity as 12.0 M
- Final volume is 1.20 L
We are required to determine the final molarity of the solution;
- Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
- Using dilution formula we can determine the final molarity.
M1V1 = M2V2
M2 = M1V1 ÷ V2
= (12.0 M × 0.035 L) ÷ 1.2 L
= 0.35 M
Thus, the final concentration of the solution is 0.35 M
Answer:
B: +3
Explanation:
If Gallium loses 3 electrons, it will become an ion.
The ion will be positively charged because in this new ion, the number of electrons is lesser than the number of protons. The charge difference will impart a positive net charge on the ion.
- In a neutral atom, the number of electrons and protons are the same.
- For positively charged ions, the number of protons is greater than the electrons
If Gallium the loss of 3 electrons offsets the charge balance in the chemical specie. Thus, the ion will have a net +3 charge.
Answer:
similarities: they both carry a charge
differences:
polyatomic ion- composed of more than one atom
monatomic ion- composed in a single atom.
Explanation:
The given precipitation reaction will be as follows.

Here, AgCl is the precipitate which is formed.
It is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.
Molarity = 
0.269 M = 
no. of moles = 0.306 mol
As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.
No. of moles = 
0.307 mol = 
mass = 43.99 g
Thus, we can conclude that mass of precipitate produced is 43.99 g.