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Eddi Din [679]
2 years ago
13

The concentration of oxalate ion (C2O42-) in a sample can be determined by titration with a solution of permanganate ion (MnO4-)

of known concentration. The net ionic equation for this reaction is 2MnO4- + 5C2O42- + 16H+ --> 2Mn2+ + 8H2O + 10CO2 A 30.00 mL sample of an oxalate solution is found to react completely with 21.93 mL of a 0.1725 M solution of MnO4-. What is the oxalate ion concentration in the sample?
Chemistry
1 answer:
zmey [24]2 years ago
6 0

Answer:

0.3152 M

Explanation:

The reaction that takes place is:

  • 2MnO₄⁻ + 5C₂O₄⁻² + 16H⁺ → 2Mn⁺² + 8H₂O + 10CO₂

First we <u>calculate the MnO₄⁻ moles used up in the titration</u>, <em>by multiplying the volume times the concentration</em>:

  • 21.93 mL * 0.1725 M = 3.783 mmol MnO₄⁻

Then we <u>convert MnO₄⁻ moles to C₂O₄⁻² moles</u>:

  • 3.783 mmoles MnO₄⁻ * \frac{5mmolC_2O_4}{2mmolMnO_4}= 9.457 mmol C₂O₄⁻²

Finally we <u>calculate the oxalate ion concentration</u>,<em> by dividing the moles by the volume</em>:

  • 9.457 mmol C₂O₄⁻² / 30.00 mL = 0.3152 M
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In the gaseous state, chlorine exists as a diatomic molecule Cl2 (Molar mass = 70.9 g/mol). Calculate the number of moles of chl
Annette [7]

Answer:

3.67 mol Cl

Explanation:

We need to convert g of Cl 2 to moles of Cl. First we divide 130 gCl2  by the molar mass (70.90 gCl2/mol) to find out how many moles of Cl2 do we have.

130 gCl2 x \frac{1 mol Cl2 }{70.90 gCl2} = 1.83 mol Cl2

Then we need to convert 1.83 mol de Cl2 to moles of Cl. We have 2 moles of Cl in every Cl2 molecule so we just need to multiply by 2.

1.83 molCl2 x \frac{2 molCl}{1 molCl2} = 3.67 molCl

8 0
3 years ago
Read 2 more answers
How many moles of carbonate are there in sodium carbonate​
vaieri [72.5K]

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>
  • The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

  • no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

  • Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

  • = 46 + 12 + 48 = 106g/mol

  • mass of CO3 = 12 + 48 = 60g

  • no. of moles of CO3 = 60/106

  • no. of moles of CO3 = 0.566mol

  • Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

5 0
2 years ago
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A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a
Andrew [12]

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

  • Free fall equation: Vf = Vo - gt

  • Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

  • Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

  • 0 = Vo - gt

   

  • Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0
3 years ago
What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?
asambeis [7]
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis. 

Hope this would help~ </span>
6 0
3 years ago
How many moles of HNO3 are present in 450 g of HNO3?
iVinArrow [24]
Since there are 63.01284 grams to one mole of HNO3, then 12.5 moles would be in 450 grams of it.
6 0
3 years ago
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