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DaniilM [7]
4 years ago
5

What mass of propane is necessary to react with the amount of oxygen in the chemical formula C3H8 + 5O2 → 3CO2 + 4H2O

Chemistry
1 answer:
Aleks [24]4 years ago
6 0
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

n(O₂)=1.407 mol
M(C₃H₈)=44.1 g/mol

m(C₃H₈)/M(C₃H₈)=n(O₂)/5

m(C₃H₈)=M(C₃H₈)n(O₂)/5

m(C₃H₈)=44.1*1.407/5=12.410 g

the mass of propane is 12.410 g
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slava [35]

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Explanation: In cobalt-silver voltaic cell, one half of the cell consists of cobalt electrode immersed in Co(NO_3)_3 solution ( which means that Co^{3+} are present in the solution) and other half of the cell consists of the Ag electrode immersed in AgNO_3 solution ( which means that Ag^+ is present in the solution)

The two electrodes are joined by the copper wire. The cobalt electrode acts as an anode and the silver electrode acts a  cathode.

At anode, oxidation reaction takes place and at cathode, reduction reaction takes place.

At Anode :                    Co(s)\rightarrow Co^{3+}(aq.)+3e^-

At Cathode:                   [Ag^+(aq.)+e^-\rightarrow Ag(s)]\times 3

Net ionic equation:   Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)

6 0
3 years ago
Use the balanced equation given below to solve the problem that follows: Calculate the mass in grams of water produced along wit
saveliy_v [14]

Answer: 1.98 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}  

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The balanced given equation is:

C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)  

According to stoichiometry :

4 moles of CO_2 will produce =  2 moles of H_2O

Thus 0.22 moles of CO_2 will produce=\frac{2}{4}\times 0.22=0.11moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.11moles\times 18g/mol=1.98g

Thus 1.98 g of water is produced along with 5.0 L of CO_2 at STP

6 0
3 years ago
Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit
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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.

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4 0
3 years ago
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Answer:

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Explanation:

3 0
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In a molecular formula of lithium sulfide there are:

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