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IRISSAK [1]
3 years ago
8

_Li + __F2 → _LIF How do I balance this?

Chemistry
1 answer:
Semmy [17]3 years ago
5 0

Answer:

 2Li   +   F₂    →    2LiF

Explanation:

The reaction expression is given as:

          Li   +   F₂    →     LiF

We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.

 Let use a mathematical approach to solve this problem;

  Assign variables a,b and c as the coefficients that will balance the expression:

             aLi   +  bF₂    →     cLiF

Conserving Li: a  = c

                   F:  2b  = c

          let a = 1, c  = 1 and b  = \frac{1}{2}  

  Multiply through by 2;

        a  = 2, b = 1 and c  = 2

               2Li   +   F₂    →    2LiF

 

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posledela
The formula for the monoprotic acid is taken as HA, reaction with base is as follows;
HA + NaOH ---> NaA + H₂O
Stoichiometry of acid to base is 1:1
At the neutralisation point, number of HA moles = number of base moles
Number of NaOH moles reacted = 0.100M / 1000 mL /L x 30.0 mL = 0.003 mol
Therefore number of HA moles reacted = 0.003 mol
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3 years ago
6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/
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6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

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Beer's Law is

A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}

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A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}

7) Absorbance

A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}

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3 years ago
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8 0
3 years ago
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