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kozerog [31]
3 years ago
12

CH3-CH2-CH=CH-CH=CH2

Chemistry
1 answer:
o-na [289]3 years ago
8 0

Answer:

Hydration (of an alkene)

Mechanism : Electrophilic addition.

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alukav5142 [94]
I can help you! What is your question?
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3 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
3 years ago
What are 4 other gases in the atmosphere besides oxegen and nitergin?
Levart [38]
Helium, neon,nitrogen and argon
5 0
3 years ago
Aluminum oxide reacts with hydrochloric acid to produce salt. What is the name of the salt produced?
KIM [24]

Answer:

IT IS CHLORIDE!

4 0
3 years ago
A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5
Marina CMI [18]

Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

1.Use Boyle's Law(P_{1} V_{1}= P_{2} V_{2}). Re-arrange to solve for V_{2}<em> for the final volume.</em>

<em />

<em>2. Plug in values. </em>V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)}  = 7.14 mL

3 0
2 years ago
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