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allsm [11]
2 years ago
14

For the following acids of varying concentrations, which are titrated with 0.50 M NaOH, rank the acids in order of least to most

volume of base needed to completely neutralize the acid.
a. 0.2M H2C6H5O7
b. 0.2M H2C2O4
Chemistry
1 answer:
k0ka [10]2 years ago
3 0

Answer:

0.2M H2C6H5O7 < 0.2M H2C2O4

Explanation:

A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.

More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.

Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.

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A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma
Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

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2 years ago
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If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration
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2 years ago
6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13
vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as C_7 H_6 O_3</em>

<em></em>

<u>Explanation:</u>

Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$

mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$

So the empirical formula is C_7 H_6 O_3

Empirical formula mass

=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

=1 \times C_7 H_6 O_3

Compound X  = C_7 H_6 O_3  is the Answer

8 0
2 years ago
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