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ipn [44]
3 years ago
14

if lithium loses an electron to become li+ what is the average atomic mass of the lithium ion? explain how you arrived at your a

nswer
Chemistry
2 answers:
Pavlova-9 [17]3 years ago
8 0
The mass of an electron 1/1823 amu's = 0.0005 amu's... the average mass of litium is 6.941 amu's. by losing an electron, the mass of average atomic mass of the Lithium doesn't really change much. 
Aliun [14]3 years ago
4 0

<u>Answer:</u> The average atomic mass of the lithium ion remains the same as lithium atom which is 6.94 amu

<u>Explanation:</u>

Mass number is defined as the sum of number of protons and number of neutrons present in an atom. It is represented as A.

A = Mass number = Number of neutrons + Number of protons.

An ion is formed when a neutral atom looses or gains electrons.

  • When an atom looses electrons, it results in the formation of positive ion known as cation.
  • When an atom gains electrons, it results in the formation of negative ion known as anion.

So, the loss or gain of electrons does not effect the mass number of an atom. Thus, when lithium atom looses an electron to form lithium ion (Li^+), the average atomic mass of the ion remains the same as the neutral atom.

Average atomic mass of lithium atom = 6.94 amu

Hence, the average atomic mass of lithium ion is 6.94 amu.

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Answer:

Gibbs equation helps us to predict the spontaneity of reaction on the basis of enthalpy and entropy values directly. When the reaction is exothermic, enthalpy of the system is negative making Gibbs free energy negative. Hence, we can say that all exothermic reactions are spontaneous.

5 0
3 years ago
Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
Elden [556K]

<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

  • <u>For magnesium chloride:</u>

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

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