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tangare [24]
4 years ago
8

Based on its electron configuration which element will most likely gain electrons from another element when forming an ionic com

pound? Potassium , Vanadium, Iodine, Xenon
Chemistry
1 answer:
Setler79 [48]4 years ago
5 0
Iodine is going to form anions capturing electrons from some other element more electropositive

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An atom has 9 electrons and 9 protons at the start. If it loses 2 electrons, the net charge on the atom will be . If the atom in
lara31 [8.8K]
Loses two = +2
gains 4 = -4
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3 years ago
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What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 40.88 mL of a 0.3842 M solution o
tatyana61 [14]

Explanation:

As molarity is the number of moles placed in a liter of solution. Therefore, no. of mole = Molarity × volume of solution in liter

Hence, moles of Cu(NO_{3})_{2} will be calculated as follows.

No. of mole of Cu(NO_{3})_{2} = 0.3842 M \times 0.04388 L = 0.0168 mole

According to the given reaction, 2 mole of Cu(NO_{3})_{2} react with 4 mole of KI.

Therefore, for 0.0168 mole amount of Cu(NO_{3})_{2} required will be as follows.

Cu(NO_{3})_{2} = 0.0168 \times \frac{4}{2}

                       = 0.0337 mole of KI

Hence, volume of KI required will be calculated as follows.

               Volume = \frac{\text{no. of moles}}{Molarity}

    Volume of KI = \frac{0.0337}{0.2089}

                           = 0.1614 liter

                            = 161.4 ml            (as 1 L = 1000 mL)

Thus, we can conclude that 161.4 ml  volume of a 0.2089 M KI is required for the given situation.

5 0
4 years ago
The atomic symbols of nitrogen and oxygen are N and O, respectively. The chemical formula of a particular substance is NO2. Whic
muminat
A compound made out of 1 nitrogen atom and 2 oxygen atoms
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3 years ago
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Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta
Leokris [45]

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

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4 years ago
Please help!!! I’ll give brainlist
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