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3 years ago

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A chemist prepares a solution of sodium thiosulfate by measuring out of sodium thiosulfate into a volumetric flask and filling t

he flask to the mark with water. calculate the concentration in of the chemist's sodium thiosulfate solution. round your answer to significant digits.

1 answer:

marshall27 [118]3 years ago

4 0

Mass of sodium thiosulfate $Na_{2}S_{2}O_{3}$ is 110. g

Volume of the solution is 350. mL

Calculating the moles of sodium thiosulfate:

$110. g Na_{2}S_{2}O_{3} * \frac{1 mol Na_{2}S_{2}O_{3}}{158.1 g Na_{2}S_{2}O_{3}}$ = 0.696 mol $Na_{2}S_{2}O_{3}$

Converting the volume of solution to L:

$350. mL * \frac{1 L}{1000 mL} = 0.350 L$

Finding out the concentration of solution in molarity:

$\frac{0.696 mol}{0.350 L} = 1.99 mol/L$

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$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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$V_1$ = initial volume of $O_2$ gas = 280 mL

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Now put all the given values in the above equation, we get:

$\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}$

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How many grams are equivalent to 1.80x10-4 tons?

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There are three isotopes of X element (X):

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Average atomic mass = 17.5 amu.

Explanation:

Given data:

X-17 isotope = atomic mass17.2 amu, abundance:78.99%

X-18isotope = atomic mass 18.1 amu, abundance 10.00%

X-19isotope = atomic mass:19.1 amu, abundance: 11.01%

Average atomic mass of X = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100

Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100

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3 years ago

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

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The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, 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 stomach acid which is HCl

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

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

Putting values in above equation, we get:

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Hence, the volume of HCl neutralized is 1.25 mL

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