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Natalija [7]
4 years ago
10

Help me with this question!!

Chemistry
1 answer:
skelet666 [1.2K]4 years ago
8 0
Nitrogen is ...19.7
flourine is 80.3...
hope this helps!
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Draw the compound that would produce 4-ethyl-3-hexanol in the presence of a nickel catalyst and hydrogen.
MatroZZZ [7]
In organic chemistry, there are already known reaction for synthesis that have been studied and experimented by scientists. For reactions involving a Nickel catalyst in the presence of hydrogen gas, the reaction would be hydrogenation of alkene to yield an alcohol. These reaction is anti-Markovnikov, or proceeds by violating the rule of Markovnikov. His rule states that 'the rich gets richer'. This means that the heavily substituted C atom of the alkene chain would receive another substituent. Therefore, the complete reaction would be

4-ethyl-3-hexene   ------->  4-ethyl-3-hexanol

In structural formula, the reaction goes as follows as shown in the picture.

5 0
3 years ago
Read 2 more answers
The temperature of 500 mL of water is 16°C what happens to the water when thermal energy is transferred to it
Lunna [17]

Answer:

The temperature of the water increases because the average

kinetic energy of the water molecules increases.Explanation:

3 0
3 years ago
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
3 years ago
Are two atoms of the same element identical??​
Neko [114]
No. Although two such atoms are essentially chemically identical (they will chemically react in the same way), they are not completely identical.
3 0
3 years ago
Use the ΔHrxn values of the following reactions: 2SO2(g) + O2(g) → 2SO3(g) ΔHrxn = –196 kJ 2S(s) + 3O2(g) → 2SO3(g) ΔHrxn = –790
sesenic [268]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -297 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

S(s)+O_2(g)\rightarrow SO_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)    \Delta H_1=-196kJ

(2) 2S(g)+3O_2(g)\rightarrow 2SO_3(g)     \Delta H_2=-790kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=\frac{[1\times (-\Delta H_1)]+[1\times \Delta H_2]}{2}

Putting values in above equation, we get:

\Delta H^o_{rxn}=\frac{[(1\times -(-196))+(1\times (-790))}{2}=-297kJ

Hence, the \Delta H^o_{rxn} for the reaction is -297 kJ.

8 0
3 years ago
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