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larisa86 [58]
3 years ago
14

Draw the major monobromination product formed by heating the following alkane with bromine.

Chemistry
1 answer:
yawa3891 [41]3 years ago
8 0
A monobromination reaction of an alkane involves an alkane and bromine. The position of the hydrogen atom that will be substituted by the bromine free radical will depend on the order of the alkane. The bromine will attach to the carbon that has the most substituents.
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Convert 281 K to degrees Celsius.<br><br> 281 K = _____<br><br> -8°C<br> 554°C<br> 8°C
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8°C

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The rising magma that may result from this type of plate movement may produce<br><br> HELP
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7 0
2 years ago
Select True or False from the pull down menus for the following statements.
Shkiper50 [21]

Answer:

First question: 1- False, 2- True, 3- False, 4 -Tue, 5- True.

Second question: 12.7 ºC

Explanation:

First question:

1- When a phase change for a pure substance is taking place under constant pressure, the temperature remains constant, and there's no sensitive heat flowing, but there's latent heat flow, which must be added to separate the molecules and to increase the kinetic energy.

2- When observed the heating and the cooling curve, at the phase change, there is no change in temperature, so it must be a horizontal line, which has a slope equal to 0.

3- Heat is the energy that is transferred by the substances or bodies because of a difference in temperature. The temperature is the measure of average kinetic energy in the molecules, so they are different.

4- As explained above, it's true, that's the definition of temperature.

5- Melting and freezing are the opposite processes and they occur at the same temperature. The difference is that for melting, the substance is absorbing heat, and for freezing it is losing heat, but the heat amount is the same for both process and is calculated by Q = ±m*L, where Q is the heat, m the mass, L the heat capacity, and the signal ± indicates if the substance is absorbing (+) or losing (-) heat.

Second question:

For the conservation of energy, the total amount of heat must be 0. The coin is losing heat, so it must be negative. The water is gaining heat, so it must be positive:

Qw - Qc = 0

Q = m*s*ΔT, where Q is the heat, m is the mass, s is the specif heat, and ΔT the temperature variation (final - initial). Qw is from water and Qc for the coin. The specif heat from the water is 4.184 J/gºC. At the thermal equilibrium, the final temperature must be equal for both.

mw*sw*ΔTw = mc*sc*ΔTc, if the coin is pure silver, sc = 0.233 J/gºC

27.0*4.184*(T - 15.5) = 15.5*0.233*(T - 100)

112.968*(T - 15.5) = 3.6115*(T - 100)

112.968T - 1751.004 = 3.6115T - 361.15

109.3565T = 1389.854

T = 1389.854/109.3565

T = 12.7ºC

So, the final water temperature would be 12.7ºC, which is impossible because it needs to increase. So the coin is not silver pure.

7 0
3 years ago
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