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GenaCL600 [577]
3 years ago
13

A sample of 17.0 M concentrated H2SO4 stock solution with a volume of 25.0 cm3 was diluted to final concentration of 5.0 M H2SO4

solution. Calculate, what is the amount of water that would be required to dilute the concentrated H2SO4 solution to a final concentration of 5.0 M?
Chemistry
1 answer:
Art [367]3 years ago
6 0

Answer:

60 cm³ of water

Explanation:

We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:

Concentration of stock solution (C₁) = 17 M

Volume of stock solution (V₁) = 25 cm³

Concentration of diluted solution (C₂) = 5 M

Volume of diluted solution (V₂) =?

C₁V₁ = C₂V₂

17 × 25 = 5 × V₂

425 = 5 × V₂

Divide both side by 5

V₂ = 425 / 5

V₂ = 85 cm³

Thus, the volume of the diluted solution is 85 cm³

Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:

Volume of stock solution (V₁) = 25 cm³

Volume of diluted solution (V₂) = 85 cm³

Volume of water =?

Volume of water = V₂ – V₁

Volume of water = 85 – 25

Volume of water = 60 cm³

Therefore, 60 cm³ of water is needed to dilute the solution.

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Your lab partner named this compound 3-methyl-4-n-propylhexane, but that is not correct.
loris [4]

<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

<u>Explanation:</u>

The IUPAC nomenclature of alkanes are given as follows:

  • Select the longest possible carbon chain.
  • For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
  • A suffix '-ane' is added at the end of the name.
  • If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

We are given:

An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms

The image of the given alkane is shown in the image below.

Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

4 0
3 years ago
Consider the compound mgcl2 what type of compound does it represent?
Naily [24]
MgCl2 is a ionic compound because it is formed by a metal (Mg) and a non-metal (Cl)
4 0
3 years ago
An electrochemical cell is constructed such that on one side a pure lead electrode is in contact with a solution containing Pb2+
Ugo [173]

Answer:

-490.7 K

Explanation:

Given:

[Ni^2+]= 0.4 M

[Pb^2+]=0.002 M

∆V= -0.012 V

VNi= -0.250V

VPb= -0.126V

F= 96500 C

R= 8.314 JK-1 mol-1

n= 2

From

T= -nF/R [∆V-(VNi-VPb)/ln [Pb2+]/[Ni2+]]

T= 2(96500)/8.314[ (-0.012) -(-0.250) - (-0.126))/ln[0.002]/[0.4]

T= 23213.856(0.112/(-5.298))

T= -490.7 K

4 0
3 years ago
1. What role do you think popular belief has on scientific discoveries?
raketka [301]

J fnjoq j wjqs fhdfwx

Its a sercret code figure it out and the answer is yours

7 0
3 years ago
2) C_H,(g) + 30,(g) → 2 CO2(g) + 2 H2O(g)
nalin [4]

Answer:

a. 2.7 mol of water

b CH2.

c. O2

Explanation:

The complete equation of the reaction should be:

2CH2(g) + 3O2(g) → 2 CO2(g) + 2 H2O(g)

a) how many moles of water will be  made?

To make 2 molecules of water (H2O) we need 2 molecules of CH and 3 molecules of O2.

We have 2.7 mol of CH2, the possible yield of water produced if it all used up will be:

2.7 mol * 2/2= 2.7 mol

We have 6.3 mol of O2, the possible yield of water produced  if it all used up will be:

6.3 mol * 2/3 = 4.2 mol

Since the maximum yield of CH2 lower, we can have 2.7 mol of water and have some excess oxygen at the end of the reaction.

b) What is the limiting reactant?

A limiting reactant is a reactant that will be used up in the reaction. This reactant has the lowest stoichiometric ratio compared to other reactants, which make them the one depleted out first. Since they depleted, the reaction will stop. Thus they limit the number of reactions and called limiting reactants. If you add the limiting reactant, the reaction will continue.

The limiting reactant in this reaction is the CH2. When producing water molecules, all 2.7 mol of CH2 will be used while we still have O2 left.

c) What is the excess reactant?

The excess reactant will have some remains after the reaction stop. That is because the excess reactant has more mass than needed for the reaction that will use all limiting reactants. Since we still have remains, adding excess reactant won't continue the reaction.

The excess reactant in this question is O2 since it still has remained after we make 2.7 mol of water. The O2 remaining, in this case, will be:

6.3 mol - 2.7mol * 3/2= 2.25 moles

8 0
3 years ago
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