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1 year ago

Below is a short reaction sequence. Provide the structure of the two missing organic compounds.

Chemistry

1 answer:

dimaraw [331]1 year ago

4 0

The product of this reaction is a halohydrin as shown here.

<h3>What are the products?</h3>

We have a reaction that first involves the formation an alkene as the bases are used on the first substrate. The alkene that is thus created is now able to react with the bromine in water.

The first step of the reaction is where the multiple bond is created and this multiple bond is what can now go on to participate in a chemical reaction in the next step of the process of reaction as shown in the image.

Bromine in water is also hat we call bromine water. This bromine water is able to add across a double bond and when that happen we will have a saturated compound. This could also be regarded as halohydrin reaction.

Recall that the first step of the reaction of the alkene with bromine is the formation of a dibromide via the brominium intermediate. This is now followed by reaction with water to form the halohydrin product.

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Write and balance the following equation:

Maru [420]

Answer:

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3 years ago

Explain the greenhouse gas effect

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“The trapping of the suns warmth in a planets lower atmosphere, due to the greater transparency of the atmosphere to visible radiation from the sun than to infrared radiation emitted from the planets surface.”

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3 years ago

3. give reasons

soldi70 [24.7K]

Answer:

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Explanation:

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3 0

2 years ago

a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp

Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH = 11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity = 4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

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4 0

1 year ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

2 years ago