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Strike441 [17]
3 years ago
7

If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate

suspended in 11 g of water, what will be the mass ratio of the materials?
Chemistry
1 answer:
Phantasy [73]3 years ago
3 0

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:    

\frac{Mass of glue}{Mass of glue + Mass of water}  

\frac{10 g}{10 g + 13 g }  

\frac{10 g}{23 g }

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}  

\frac{ 8 g}{ 8 g + 11 g }  

\frac{8 g}{19 g }

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Explanation:

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How many moles of water are produced when 3.45 moles of KMnO4 react? Give your answer to the nearest 0.1 moles. moles H2O
andrew11 [14]

Answer:

13.8 moles of water produced.

Explanation:

Given data:

Moles of KMnO₄ = 3.45 mol

Moles of water = ?

Solution:

Chemical equation:

16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Moler ratio of water and KMnO₄:

             KMnO₄       :        H₂O

                2             :          8

               3.45         :         8/2×3.45 = 13.8 mol

Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.

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Which is a type of star system?
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A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your a
Molodets [167]

Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

Mass of CH_3C0_2 H= 77 \ g \\\\

Molar mass of CH_3C0_2 H = 60.05 \  \frac{g}{mol} \\\\

No of moles in n_{CH_3CO_2H} = 77 \ g \times  \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}  

                                          = 1.28 \ mol \ CH_3CO_2H

Mass of water (H_2O)= 42 \ g

The molar mass of water (H_2O)= 18.02 \  \frac{g}{mol}

moles of water n_{H_2O}:

= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O

Molfraction of acetic acid:

X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\

                     =\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35

7 0
3 years ago
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