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r-ruslan [8.4K]
3 years ago
12

How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?

Chemistry
1 answer:
Allushta [10]3 years ago
3 0
I dont know but do you know da wae brudda?


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Consider the titration of 10.00 mL of a monoprotic weak acid with 0.1234 M NaOH. If the equivalence point volume of NaOH was det
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The initial concentration of the unknown acid is 0.1900 M.

Explanation:

Titration is a chemical method of analysis to know the concentration and volume of the unknown chemical or analyte.

The formula for the titration is:

Macid x Vacid = Mbase x V base

The volume must be in litres. The volume is given in ml it should be divided with 1000 to obtain values in litre.

Data given are:

volume of acid= 10 ml 0.01 L

Molarity of the acid = ?

volume of the NaOH or base = 15.4 ml or 0.0154 L (equivalence point of the base)

molarity of the base = 0.1234 M

Applying the formula and putting the values, we get

Macid x 0.01 = 0.1234 x 0.0154

Macid =  0.1900 M

The weak acid is having molarity of 0.1900 M against the strong base with molarity of 0.1234M.

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