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Sergio039 [100]
3 years ago
5

2) Consider the reaction of 1-bromopropane with a solution of sodium ethanoate to form

Chemistry
1 answer:
fredd [130]3 years ago
3 0

Answer:

(a). <em>CH3CH2CH2Br + CH3COONa ===> CH3COOCH2CH2CH3 + </em><em>NaBr</em><em> </em>

<em>-propylethanoate</em><em> </em>or <em>ethylpropanoate</em>

Another organic product is <em>CH3COOH</em><em> </em><em>(</em><em> </em><em>ethanoic</em><em> </em><em>acid</em><em> </em><em>)</em><em>.</em>

It will be in smaller amounts than the ether

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Answer:

Neutrons or Neutral atoms

Explanation:

Neutral atoms of each element contain an equal number of protons and electrons. The number of protons determines an element's atomic number and is used to distinguish one element from another.

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Why is it called endogenous process
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7 0
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If a solution conducts electricity and has a pH value of 7 then it could be
stealth61 [152]
The solution that conducts electricity and has a pH value of 7 would most likely be a neutral solution. Water is among the best examples of a neutral solution. When the pH of a solution is considered to be lesser than 7, it is an acid, while if the pH is greater than 7, it is considered to be a basic solution.
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Which of the following is the best way to separate sugar from water?
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4 0
3 years ago
Read 2 more answers
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
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