By whom is chemistry used in the real world

Fritz-Haber process

maks197457 [2]

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

6 0

3 years ago

What are the major organic products are formed when the following compounds react with methylmagnesium bromide (CH3MgBr), follow

Sliva [168]

Answer:

Major organic products are- (a) propan-1-ol and (b) 2-methylpropan-2-ol

Explanation:

methyl magnesium bromide gives nucleophilic addition reaction with carbonyl group. Because methyl magnesium bromide is a strong nucleophile and carbonyl group is a strong electrophilic center.

Propanal contains an aldehyde group and propanone contains a ketone group. hence they both give nucleophilic addition with methyl magnesium bromide.

Dilute acid is added to protonate the alkoxide produced during nucleophilic addition.

Reactions are shown below.

6 0

3 years ago

How many meters in 1000 mm

iris [78.8K]

1000 mm equals 1 meter

7 0

4 years ago

Read 2 more answers

In covalent bonding, electrons are shared within molecules. This is because..

avanturin [10]

Answer:

You're answer is B.

Explanation:

Covalent Bonds share a negative and positive ion.

6 0

3 years ago

The rate of decomposition of N2O5 in CCl4 at 317 K has been studied by monitoring the concentration of N2O5 in the solution. 2 N

pav-90 [236]

Answer:

Average rate of reaction is 0.000565 M/min

Explanation:

Applying law of mass action for the given reaction:

Average rate = $-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}$

Where, $-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}$ represents average rate of disappearance of $N_{2}O_{5}$ , $\frac{1}{4}\frac{[NO_{2}]}{\Delta t}$ represents average rate of appearance of $NO_{2}$ and $\frac{[O_{2}]}{\Delta t}$ represents average rate of appearance of $O_{2}$

Here, $-\frac{[N_{2}O_{5}]}{\Delta t}$ = $-\frac{(2.16-2.36)}{(177-0)}M/min=0.00113M/min$

So average rate of reaction = $[tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}$ [/tex] = $\frac{1}{2}\times (0.00113M/min)=0.000565M/min$

7 0

3 years ago