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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Answer:
32.07 g/mole.
Explanation:
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Answer:
- <em>To balance a chemical equation it may be necessary to adjust the </em><u>coefficients.</u>
Explanation:
The <em>coefficients</em> of a <em>chemical equation</em> are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.
By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.
Here an example.
The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:
- Word equation: hydrogen + nitrogen → ammonia
- Skeleton equation: H₂ + N₂ → NH₃
This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.
So, in order to comply with the law of mass conservation you adjust the coefficients as follow.
- Balanced chemical equation: 3H₂ + N₂ → 2NH₃
As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:
- 3 mol H₂ : 1 mol N₂ : 2 mol NH₃
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Answer:
47.36mL
Explanation:
Using Boyles law equation, which states that:
P1V1 = P2V2
Where;
V1 = initial volume (mL)
V2 = final volume (mL)
P1 = initial pressure (atm)
P2 = final pressure (atm)
Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm
First, we need to convert 152kPa to atm by dividing by 101
1kPa = 0.0099atm
152kPa = 1.505atm
P1V1 = P2V2
1.505 × 25.3 = 0.804 × V2
38.08 = 0.804V2
V2 = 38.08/0.804
V2 = 47.36mL
