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Answer:

Explanation:
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In this case, mercury (II) oxide (HgO) is obtained via the reaction:

Nonetheless, since it is a reaction carried out in basic solution, mercury's half-reaction only, must be:

Thus, it is seen that OH ionis should be added due to the basic aqueous solution considering that 2 electrons are transferred from 0 to 2 in mercury.
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Answer:
Zero
Explanation:
FrBr is an ionic compound
.
Fr is in Group 1. Br is in Group 17.
The charges on the ions are +1 and -1, respectively.
The compound consists of Fr⁺Br⁻ ions.
However, there are equal numbers of + and - charges, so
The overall charge of the compound is zero.
Answer:
0.077 M
Explanation:
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
Those are the correct steps, young chemist. Don't be discouraged by an insane answer.