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Otrada [13]
3 years ago
12

The unsaturated hydrocarbons will react readily with other elements or compounds. True False

Chemistry
2 answers:
statuscvo [17]3 years ago
6 0
Your answer is 'true'.
Ber [7]3 years ago
6 0

The given statement is true.  

The hydrocarbons that comprise double or triple covalent bonds between the adjacent carbon atoms are known as unsaturated hydrocarbons. Those with at least one carbon-to-carbon triple bond are known as alkynes and those with at least one carbon-to-carbon double bonds are known as alkenes.  

Other than aromatic compounds, the unsaturated hydrocarbons undergo addition reactions to their multiple bonds and are very reactive. Thus, unsaturated hydrocarbons will react easily with other elements or compounds.  

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How many particles are in 7.07x10^-6 moles of a substance?
Butoxors [25]

Answer:

4.25*10^18

Explanation:

1 mole =6.023*10^23 particles

so 7.07*10^-6 mole=6.023*10^23*7.07*10^-6 particles=4.25*10^18 particles

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The physical properties of matter produce many cause and effect relationships. Which effort is produced by the close packing of
stepan [7]

Answer:

A.

Explanation:

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While the gaseous states gives the highest freedom of movement, the solid state gives the least freedom of movement. Molecules and atoms of a solid are so tightly packed.

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3 years ago
For parts a & b below, derive only the initial value problem set up.
Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )

The rate of the sugar coming into the tank is 0

\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min

So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100}  \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C  \\ \\  \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So;  y(t) = 25 e^{-\dfrac{t}{100}

Thus, the amount of sugar at any time t = \mathbf{25 e^{^{-\dfrac{t}{100}}}}

B) For tank B :

\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\  \dfrac{dy}{50-y }= \dfrac{dt}{50}

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3 years ago
Why doesn't your consist of 365 days and A day of 24 hours
LenKa [72]
Hi what do you mean by this? i dont understand at all pls explain :)
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3 years ago
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