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Vika [28.1K]
3 years ago
12

Convert ethene to propanol​

Chemistry
1 answer:
klemol [59]3 years ago
7 0

Answer:

The dehydration of ethanol Highlights Reductive Hydroformylation of ethene to propanol with homogeneous rhodium catalysts. Propane can be oxidized to propanol, and then dehydrated to form propene.

Explanation:

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A classification of matter which stipulates that the compound must contain carbon is called
Ronch [10]

Answer:c

Explanation:

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3 years ago
How many grams of NaOH are produced with the reaction of 5.00 moles of water? *
Andre45 [30]

Answer:

200 grams of NaOH are produced with the reaction of 5.00 moles of water

Explanation:

First of all you apply a rule of three to know the amount of moles of NaOH as follows: if 2 moles of water produce 2 moles of NaOH, 5 moles of water how many moles would they produce?

moles of NaOH=\frac{5 moles of water*2 moles of NaOH}{2 moles of water}

moles of NaOH= 5

Being:

  • Na: 23 g/mole
  • O: 16 g/mole
  • H: 1 g/mole

the molar mass of NaOH is: 23 g/mole + 16 g/mole + 1 g/mole= 40 g/mole

Then a rule of three applies as follows: if in 1 mole there are 40 g of NaOH, in 5 moles how much mass is there?

mass=\frac{5 moles*40g}{1  mole}

mass= 200 g

<u><em>200 grams of NaOH are produced with the reaction of 5.00 moles of water</em></u>

4 0
4 years ago
In a biology class, students are studying different liquid fertilizers. During the class, Maria knocks over a container of pesti
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A solution is made by dissolving 25.20 g of aluminum acetate in 185.0 mL of water. What is the concentration of acetate ions?
Andreas93 [3]
To answer this question, you need to know the aluminum acetate weight and formula. Molecular weight of aluminum acetate is 204.11g/mol, so the concentration should be: (25.20/204.11 )/(0.185l) = 0.1234 mol/0.185l= 0.6673 M<span>

</span>The aluminum acetate Al(C2H3O2)3 which mean for one aluminum acetate there will be 3 acetate ion. The concentration of acetate should be: 0.6673 M= 2M
4 0
3 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2
allsm [11]

Answer:

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
3 years ago
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