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Veseljchak [2.6K]
2 years ago
13

Vinegar is an aqueous solution of acetic acid, ch3cooh. a 5.00 ml sample of a particular vinegar requires 26.90 ml of 0.175 m na

oh for its titration. what is the molarity of acetic acid in the vinegar?
Chemistry
1 answer:
Fed [463]2 years ago
4 0

  The molarity  of  acetic acid in the  vinegar is  0.94 M


 <u><em> calculation</em></u>

Step 1:  write  the balanced equation between CH3COOH  + NaOH

that is CH3COOH   + NaOH  →  CH3COONa  + H2O


step 2 :  find the moles of NaOH

moles  =molarity  x volume in L

volume in liters = 26.90/1000=0.0269 l

moles = 0.175 mol /L x 0.0269 L  =0.0047  moles  of NaOH


Step 3: use the mole  ratio to find moles of CH3COOH

that is the  mole ratio of  CH3COOH: NaOH is 1:1 therefore  the moles of CH3COOH is  =0.0047  moles


Step 4:  find the  molarity  of  CH3COOH

molarity = moles/volume in liters

volume in liter = 5.00/1000 =0.005 l

molarity  is therefore=0.0047 moles/ 0.005 l = 0.94 M

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2 years ago
It take 38.70cm³ of 1.90m NaoH to neutralize 10.30cm³ of H2so4 in a battery, calculate the molar concentration of H2so4
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Answer:

M_{acid}=3.57M

Explanation:

Hello there!

In this case, since this acid-base neutralization is performed in a 1:2 mole ratio of acid to base as the former is a diprotic acid (two hydrogen ions in the molecule), we can write the following equation:

2M_{acid}V_{acid}=M_{base}V_{base}

In such a way, we can solve for the molarity of the acid, given the molarity and concentration of the NaOH base and the volume of the acid:

M_{acid}=\frac{M_{base}V_{base}}{2V_{acid}}

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2 years ago
Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.
icang [17]

Answer : 3.2 X 10^{15} g/cm^{3}

Explanation :  To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = 1.66054 X 10^{-24} g

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X  1.66054 X 10^{-24} g/amu = 1.673 X 10^{-24} g

Now, Volume = 1/6πd³ as diameter is given as 1.0 X 10^{-15} m converting it to cm will require to multiply with 100

∴ Volume  = 1/6π (1.0 X 10^{-15}mX 100 cm / 1 m)^{3}

Hence, volume =  5.236 X 10^{-40} cm^{3}

Therefore, Density = mass / volume

∴ Density =  1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}

Therefore, Density will be 3.2 X 10^{15} g/cm^{3}.

6 0
2 years ago
Read 2 more answers
If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

<u>Step 3: </u>Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]  

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Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

<u> Step 7:</u> Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

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