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4 years ago

What are the uses of methanol?

Chemistry

1 answer:

zmey [24]4 years ago

5 0

Methanol has severl uses few of which are discussed below.

<h3><u>Explanation</u>:</h3>

Methanol is the simplest alcohol with chemical formula CH3OH. Its nin drinkable because its toxic when consumed.

Methanol is used for preparing acetaldehyde, formaldehyde, butandiol, acetal resins etc which are used for production of furthur high specific chemicals with high rate of purity. Its also used widely in automobile industry.

Its used to produce acetic acid, solvent esters and acetates, terepthalic acid, etc which are used in automobiles, paint industry, ink industry etc.

Its used in preparation if zero emmision fuels. Its also used to produce antifreeze, lubricants, burning alcohol, spirit etc.

Its also used to produce different cleaners and canned heat.

Its used to dress wounds, for preparation of MMA, MTOs etc.

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If the spin of one electron in an orbital is clockwise, what is the spin of the other electron in that orbital?

umka2103 [35]

Answer:

The other electron must have anticlockwise spin.

Explanation:

According to the pauli exclusion principle, the two elecrton present in same orbital must have opposite spin.

If the one electron is clockwise the other must be in anti clockwise direction. The clockwise direction is represented by the sign +1/2 while anti clockwise direction is represented by -1/2.

According the pauli principle, the two electrons must have different fourth electronic quantum number. The electron in same orbital have same first three quantum number i.e, n=1 l=0 and ml =0 in case of first subshell.

3 0

3 years ago

At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l

Leona [35]

Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 3 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 1.40 L

$V_2$ = final volume of gas = 0.950 L

$T_1$ = initial temperature of gas = $35^oC=273+35=308K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.918atm$

Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

4 0

4 years ago

Read 2 more answers

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago

10. At 573K, NO2(g) decomposes forming NO and O2. The decomposition reaction is second order in NO2 with a rate constant of 1.1

leva [86]

Answer:

48.67 seconds

Explanation:

From;

1/[A] = kt + 1/[A]o

[A] = concentration at time t

t= time taken

k= rate constant

[A]o = initial concentration

Since [A] =[A]o - 0.75[A]o

[A] = 0.056 M - 0.042 M

[A] = 0.014 M

1/0.014 = (1.1t) + 1/0.056

71.4 - 17.86 = 1.1t

53.54 = 1.1t

t= 53.54/1.1

t= 48.67 seconds

Hence,it takes 48.67 seconds to decompose.

6 0

3 years ago

Four people weigh a standard mass of 10.00g on the same balance. Which of the following sets of readings suggests measurements t

Svetach [21]

An example of accurate but not precise would be 3 people weigh a 10g sample. the weights are 0g, 10g, & 20 grams. the scale is way off but the weights average to the right thing.
precise but not accurate would weighing a 10g sample 3 times and getting 5.5, 5.4, & 5.5. they'e all incredibly similar therefore precise but its nowhere near 10, so not accurate.
neither precise nor accurate would be 3 weights being 10, 20, &30. It averages wrong and is imprecise.

5 0

4 years ago