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Anna11 [10]
3 years ago
5

A chemist adds of a M barium chlorate solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist h

as added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
nordsb [41]3 years ago
7 0

The given question is incomplete, the complete question is:

A chemist adds 200.0 ml of a 0.52M barium chlorate (Ba(CIO3)2) solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.

Answer:

The correct answer is 32 grams.

Explanation:

Based on the given solution, the molarity of barium chlorate solution given is 0.52 M, this shows that the solution will comprise 0.52 moles in 1 L or 1000 ml of the solution.  

Therefore, in 200 ml, it will comprise 0.52/1000 × 200 moles of Ba(ClO₃)₂,  

= 0.52/1000 × 200 = 0.104 moles

The molecular mass of Ba(ClO₃)₂ is 304.23 gram per mole

So, the mass of Ba(ClO₃)₂ in 0.104 moles will be,  

= 304.23 g/mol × 0.104

= 31.639 grams or 32 grams.  

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A solution of malonic acid, H2C3H2O4 , was standardized by titration with 0.1000 M NaOH solution. If 21.17 mL of the NaOH soluti
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Answer: The molarity of the malonic acid solution is 0.08335 M

Explanation:

H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

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n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2C_3H_2O_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=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL

Putting values in above equation, we get:

2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M

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