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devlian [24]
3 years ago
9

What product is formed when 1-bromopropane reacts with ch3o−?

Chemistry
1 answer:
STatiana [176]3 years ago
4 0
When 1-Bromopropane is treated with methoxide it undergoes SN² Reaction. The product formed is Methyl Propyl Ether. This type of reaction is very important in synthesizing Asymmetric ethers. The alkoxides are oftenly generated by reacting Alcohols with sodium metal, which on elimination of Hydrogen gas produces corresponding alkoxide. The reaction of given statement is as follow,

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3 years ago
What reaction would be the chemical equation 2H2O->2H2+O2 represent? A.decomposition of water b. Formation of water c. Combus
vodka [1.7K]
It's decomposition
A combustion reaction involves carbon dioxide so it can't be that
The water is not being formed
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So it is decomposition, not just from process of elimination, but also because decomposing water means heating it.
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4 years ago
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To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o
Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

\pi  r^{2}h ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

r=\frac{d}{2}

=\frac{3.36x10^{2}}{2}

= 168 m

Volume =

=\frac{22\times 168^{2}\times 21.80}{7}

=1.93\times 10^{6} m^{3}

2.Convert ppm to g/m3 and Solve for mass of F-

1ppm = 1g/m^{3}

0.9ppm = 0.9g/m^{3}

Because both ppm and g/m3 are same quantity .

g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}

0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}

mass\ of\ F- =1.740g

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

2Na+F_{2}\rightarrow 2NaF

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g  of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

\frac {84}{38}\times 1.74

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0
3 years ago
Which of these do NOT show malleability a) Iron <br>b) Sulfur <br>c) Copper <br>d) Aluminum​
olganol [36]

Answer:Sulfur

Hope this helps!

Explanation:

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