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3 years ago

An aspirin tablet contains 325 mg acetylsalicylic acid. How many acetylsalicylic acid molecules does it contain?

1 answer:

4 0

The formula of acetylsalicylic acid is C9H8O4. And the gram formula mass of the molecule is 180 g/mol. So there is 325/180 = 1.8 mol. There is 6.02*10^23 molecules per mol. So the answer is 1.08 * 10^24 molecules.

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Calculate the concentration of hydrochloric acid if 225.0 mL of this acid is neutralized by 120.0 mL of 2.0m sodium hydroxide A.

timurjin [86]

M1V1 = M2V2
225M1 = (2)(120)
225M1 = 240
M1 = 240/225 = 1.066 M C

5 0

3 years ago

This type of triglyceride contains more than one double bond in the fatty acid carbon atoms

erastova [34]

polyunsaturated!!! Hope this helps mark this the brainliest!

4 0

3 years ago

Why an aspirin tablet works faster if it is crushed rather than swallowed whole?

givi [52]

Answer: because the tablet is already crushed and is easy for the organism to absorb instead of having to break it down

Explanation:

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3 years ago

How many atoms are present in 9.8g of H2SO4

aleksley [76]

Answer:

Explanation:

98 gH

contains1 mole of H

1 mole of H

contain=1 mole of sulphur atoms.

∴9.8 g of H

contain =0.1 mol of sulphur atom.

Thus, number of sulphur-atoms =n×N

(in 9.8 g of H

)

=0.1×6.023×10

=0.6023×10

7 0

3 years ago

4. A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an averag

PtichkaEL [24]

Answer:

C. 0.191 M

Explanation:

Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the reaction between HCl and NaOH:

$HCl~+~NaOH~->~NaCl~+~H_2O$

The "titrant" in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

$0.100~M=\frac{mol}{0.0382~L}$

$mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH$

Now, in the reaction, we have a 1:1 molar ratio between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:

$0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl$

If we want to calculate the molarity of the HCl solution we have to divide by the litters of HCl used in the experiment (20 mL= 0.02 L):

$\frac{0.0382~mol~HCl}{0.02~L}~=~0.191~M$

The concentration of the HCl solution is 0.191 M

I hope it helps!

8 0

3 years ago