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nikitadnepr [17]
3 years ago
14

How many moles contain 1.25x 10^15 molecules carbon dioxide?

Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0
Avagadros number states that 1 mol of any substance is made of 6.022 x 10²³ units.
These units could be atoms that make up an element or molecules that make up a compound. CO₂ is a compound therefore its made of CO₂ molecules.
1 mol of CO₂ contains 6.022 x 10²³ molecules of CO₂
Therefore if  6.022 x 10²³ molecules of CO₂ make up 1 mol 
then 1.25 x 10¹⁵ molecules of CO₂ make up - \frac{1.25 x 10¹⁵}{6.022 x 10²³}
Number of CO₂ moles present - 2.0756 x 10⁻⁹ mol of CO₂
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What is the mass of oxygen in 25.0 grams of potassium permanganate , KMnO4?
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Answer:

10.76 grams

Explanation:

Given that the amount of KMnO_4 is 25.0 grams.

Mass of 1 mole of KMnO_4 = 158 grams

The number of atoms in 1 mole of KMnO_4 is 4.

Mass of oxygen in 1 mole of KMnO_4 = 16\times 4 = 68 grams.

Here, 158 grams of KMnO_4 has 68 grams of oxygen

So, the amount of oxygen in 1 gram of KMnO_4 = 68/158 grams

Therefore,  the amount of oxygen in 1 gram of KMnO_4 = \frac {68}{158}\times 25 grams

=10.76 grams

Hence, 25 grams of KMnO_4 has 10.76 grams of oxygen.

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Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit
REY [17]

Answer:

Explanation:

According to this. Let's analize the possible products a, b and c.

First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.

With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.

Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.

Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.

Picture attached show the product a, b and c. Hope this can help

4 0
2 years ago
Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine
yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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