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Dafna11 [192]
3 years ago
8

What are alkenes used to make?

Chemistry
2 answers:
gregori [183]3 years ago
8 0
They are used as starting materials in the syntheses of alcohols, plastics, laquers, detergents, and fuels. The most important alkenes for the chemical industry are ethene, propene and 1,3-butadiene.
evablogger [386]3 years ago
3 0

Answer:

By far the most important use of alkenes is in the making of plastics as plastics are used in almost everything.

Explanation:

Alkenes themselves aren't used much in everyday life however Alkenes are very important to industrial synthesis as it is relatively easy to turn them into other things.

Alkenes can be turned into polymers or plastics through addition reactions and the most common ethene is turned into everything from plastic bags to bottles.

Alkenes can also be turned into alcohols. most commonly propene is used as a feedstock to produce butanol and other products useful in industry or for production

Explanation:

Can you be my friend in here

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Which represents a balanced nuclear equation?
shtirl [24]

Answer: second answer

Explanation:

The thing is that every new nuclear cycle a new element forms and reduces an electron. I may be wrong but this is the most logical and scientifically correct answer

3 0
3 years ago
Could you please Calculate the number of atom of 40K (potassium 40) in 1gram of KCl. Taking into account the isotopic abundance
Crank

Answer:

9.53*10^{17} atoms of 40K

Explanation:

You can use the molecular mass and the Avogadro´s number, in the following formula:

N_{40K=\frac{m_{KCl}}{M_{MKCl}}}*N_{Avogadro}*(IA_{40k})

where m_{KCl} is the sample mass, M_{KCl} is the molecular mass of the KCl and IA(40K) is the isotopic abundance of 40K.

Now replacing the values, you can find:

N_{40K}=\frac{1g}{74.5513\frac{g}{mol}}*6.022*10^{23}mol^{-1} *0.000118

N_{40K}=9.53*10^{17} atoms

8 0
3 years ago
The following situations are presented to a medical examiner. In which situation is the medical examiner most likely to gather m
kolezko [41]
B) 25-year-old male recovered from rapids after his empty kayak is sited by friends 
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8 0
3 years ago
A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re
sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula  -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass  = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature  10.0°C + 273 = 283 K

T₁ = initial temperature ,  40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ )  = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

6 0
3 years ago
Hi May I know how to balance this
almond37 [142]

Answer:

  2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

Explanation:

Equating coefficients, you get ...

  aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃

For Ba: 3a = d

For P: 2a = 4c

For O: 8a +2b = 10c +3d

For Si: b = d

__

Expressing everything in terms of b and c, we get ...

  d = b

  a = b/3 = 2c

From the second, b = 6c, so we have ...

  a = 2c

  b = 6c

  c = c

  d = 6c

And we can write the equation with c=1 as ...

  2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

4 0
3 years ago
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