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shusha [124]
3 years ago
13

Which of the following solutions contains the largest number of moles of dissolved particles? a.) 25 ml of sodium chloride b.) 2

5 ml of aluminum hydroxide c.) 25 ml of acetic acid d.) 25 ml of nitric acid e.) 25 ml of sugar
Chemistry
2 answers:
kirza4 [7]3 years ago
5 0

Answer:

b.) 25 ml of aluminum hydroxide

Explanation:

For this question, we have to assume that we have the <u>same concentration</u> for all the solutions, for example, <u>1 M</u>. Additionally, we have to take into account the <u>ionization reaction</u> for each species:

a) NaCl~->~Na^+~+~Cl^- <u>we have to ions</u>

b) Al(OH)_3~->~Al^+^3~+~3OH^- <u>we have fourth ions</u>

c) CH_3COOH~->~CH_3COO^-~+~H^+ <u>we have two ions</u>

d) C_6H_1_2O_6~->~C_6H_1_2O_6  <u>we have one ion</u>

If we have the same volume and the same concentration the variables that will help us to answer the question would be the n<u>umber of ions.</u> If we have <u>more ions we will have more particles dissolved</u>. Therefore the answer would be b) (<u>due to the fourth ions</u>).

I hope it helps

katen-ka-za [31]3 years ago
4 0

Answer:

b.) 25 ml of aluminum hydroxide.

Explanation:

Hello,

In this case, bu knowing the formula of each substance, we can predict the number of particles are going to be dissolved in the given solutions, thus we have:

Sodium chloride, NaCl: when it is placed in water, two ions, sodium and chloride are dissolved.

Aluminium hydroxide, Al(OH)3: when it is placed in water, four ions are dissolved, one aluminium (III) and three hydroxyl.

Acetic acid CH3COOH: when it is placed in water, and equilibrium dissolution is present, therefore, not all the acetic acid particles will be dissolved as ions, anyway, two ionic species are formed, acetate anion and hydronium.

Sugar C12H22O11: when it is placed in water, it is not properly dissociated but it is said to form hydrogen bonds in order to get dissolved in water, therefore, just one dissolved particle is formed.

In such a way, the substance having the larges number of moles of dissolved particles is aluminium hydroxide having four, by considering the same concentration for all the solutions, say 1 M or that so.

Regards.

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Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To balance it, we can simply add another sodium atom on the left. Hence:

\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

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\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}

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And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Given the initial value and the above ratios, this yields:

\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

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=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}

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Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

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\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}

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Cancel like units:

\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}

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\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}

Since the resulting value should have three significant figures:

\displaystyle = 65.2 \text{ g Al$_2$O$_3$}

So, approximately 65.2 grams of aluminum oxide is produced.

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