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Sveta_85 [38]
3 years ago
7

How many grams of sodium chloride are present in a 0.75 M solution woth volume of 500.0 milliliters?

Chemistry
1 answer:
Aneli [31]3 years ago
8 0
V = 500ml = 0,5 l
c = 0,75 M

c = n/V
n = c × V
n = 0,75 × 0,5 = 0,375[mol]

M NaCl = 58,5 g/mol

1 mol ------- 58,5 g
0,375 mol ----- x g

x = 0,375 mol × 58,5 g / 1 mol = 21,94 g

Answer: In this solution there are 21,94 g of sodium chloride.

:-) ;-)

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Answer:

C2H3O3

Explanation:

Empirical formula is the simplest whole number ratio of moles of atoms that you can find in a molecule.

In combustion analysis all Carbon reacts producing CO2 and all hydrogen reacts producing H2O. With the differences in masses we can find the mass of oxygen and their moles:

<em>Moles CO2 = Moles C:</em>

14.08g * (1mol/44.01g) = 0.3199 moles C * (12.01g/mol) = 3.8423g C

<em>Moles H2O:</em>

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<em>Mass O:</em>

12.01g = Mass O + 3.8423g C + 0.4797g H

Mass O = 7.688g O

<em>Moles O:</em>

7.688g O * (1mol/16g) = 0.48 moles O

The ratio of atoms (Dividing in the moles of C that are the lower number of moles):

O: 0.48moles O / 0.3199 moles C = 1.50

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H: 0.4797 moles H / 0.3199 moles C = 1.50

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O: 1.50* 2 = 3

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The empirical formula is:

<h3>C2H3O3</h3>

7 0
3 years ago
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ivanzaharov [21]

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Rate law: k[C_4H_6]^2

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The integrated rate law for second order is \frac{1}{[A]}=\frac{1}{[A]_0} +kt where A is C_4H_6.

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