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-BARSIC- [3]
3 years ago
12

A student calculated the molarity of a solution prepared by dissolving 0.730 mol of table sugar (sucrose, C12H22O11) in 1.8x10^3

mL of water as 4.06x10^-4 M C12H22O11.
Explain the student's calculation error and explain how the student should solve for the correct value of molarity. Show a valid calculation for the molarity.

II. The student then takes a 1.00 M stock solution of table sugar (sucrose, C12H22O11) and mixes 0.305 L of stock solution with additional distilled water to create a dilute solution with a total volume of 1.25 L.

Explain how the student can determine the molarity of the resulting solution. Show a valid calculation for the final molarity. PLS HELP I'M TAKING MY EXAM NOW!!!!!
Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer: C= 0.406 M

Explanation:

Solution.

ν

=

0.730

m

o

l

;

ν=0.730mol;

V

=

1.8

⋅

1

0

3

m

L

=

1.8

L

;

V=1.8⋅10

3 mL=1.8L;

C=0.730mol

1.8 L=0.406 M

C= 1.8L

0.730mol =0.406M

The student made a mistake because he did not convert a unit of volume from milliliters to liters. After all, molarity is defined as the number of moles of solute per liter of solution.

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3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?
Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

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$_2$}

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

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

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}

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$_2$}}

Cancel like units:

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

Multiply. Hence:

=0.3463...\text{ g H$_2$}

Since we should have two significant values:

=0.35\text{ g H$_2$}

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

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:

\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

\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$_3$}}

Using the given value and the above ratios, we acquire:

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

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}

Multiply:

\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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2 years ago
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What are some common tools scientists use to measure length and mass?
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To measure length scientists may use rulers, meter sticks, etc. and to measure mass they may use a balance.
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3 years ago
What changed about the peppered moth population during the second industrial revolution
Marysya12 [62]

Answer:

The months became darker colored and more darker colored moths were sighted.

Explanation:

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7 0
3 years ago
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2. A solution is made by dissolving 3.88 g of NaCl in enough water to make 67.8 mL of solution. What is the concentration of sod
Sauron [17]

Answer:

is 23

Explanation:

i got it rigjt .......

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2 years ago
If you start with 10ml of 0.75 m cu(no3)2 how much cu (s) in grams should be recovered in step #7
LenaWriter [7]
<span>0.48 grams. Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced. To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So 0.0075 mol * 63.546 g/mol = 0.476595 g. Round the results to 2 significant figures, giving 0.48 grams.</span>
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