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julsineya [31]
1 year ago
10

Write a balanced chemical equation for the dissociation of lithium bromide in water.

Chemistry
1 answer:
Novay_Z [31]1 year ago
4 0

The balanced chemical equation is :

LiBr (aq) + H2O (l)  →  LiOH (aq) + HBr (aq) .

<h3>Explain the balanced chemical equation?</h3>
  • To balance a chemical equation, add coefficients to the symbols or formulas as needed so that the same number of each type of atom occurs in both reactants and products.
  • There is a strategy that can help you balance equations faster. It is known as balancing by inspection. Essentially, you take the number of atoms on each side of the equation and add coefficients to the molecules to balance out the number of atoms.
  • Lithium bromide, abbreviated LiBr, is a chemical compound. It is water, alcohol, and ether soluble. It is made by reacting hydrobromic acid with lithium hydroxide.

To learn more about balancing chemical equation refer to :

brainly.com/question/26694427

#SPJ4

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146,1g \ \ \ \rightarrow \ \ \ \ 6,02*10^{23}\\&#10;0,65g \ \ \ \ \rightarrow \ N_{x}\\\\&#10;N_{x}=\frac{0,65g*6,02*10^{23}}{146,1g}\approx0,0268*10^{23}=2,68*10^{21}

2nd --> <span>How many grams of NH3 are needed to provide 2,68</span>ˣ10²¹ molecules


17,03g \ \ \ \ \rightarrow \ \ \ \ 6,02*10^{23}\\&#10;m_{x} \ \ \ \ \ \ \ \ \ \rightarrow \ \ \ \ 2,68*10^{21}\\\\&#10;m_{x}=\frac{2,68*10^{21}*17,03g}{6,02*10^{23}}\approx7,58*10^{-2}=0,0758g
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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
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To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

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