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Nezavi [6.7K]
2 years ago
4

How does the boiling process at supercritical pressures differ from the boiling process at subcritical pressures?

Chemistry
1 answer:
Genrish500 [490]2 years ago
4 0
Super critical, the fluid won't bubble. As you warm it, It will gradually change its properties to be an ever increasing number of gas like until the point when it winds up unclear from a gas. There is no stage change over the critical point. Though at subcritical pressures, there is dependably an distinct surface between the phases.
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50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in th
statuscvo [17]

Answer:

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.

3 0
2 years ago
Read 2 more answers
Which statement is TRUE regarding the macroscopic and
damaskus [11]

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

7 0
2 years ago
a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre
zhannawk [14.2K]
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!
3 0
2 years ago
A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl
ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)

In this case, the cathode and anode both are same. So, E^o_{cell} is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 1

E_{cell} = ?

[Cl^{-}{diluted}] = 0.0222 M

[Cl^{-}{concentrated}] = 2.22 M

Now put all the given values in the above equation, we get:

E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}

E_{cell}=0.118V

Therefore, the cell emf for this cell is 0.118 V

4 0
2 years ago
Examine the block from the periodic table for the element nitrogen, chemical symbol N. Using your knowledge of the periodic tabl
vodka [1.7K]
I believe that the answer is A) All Nitrogen atoms have 7 protons
4 0
2 years ago
Read 2 more answers
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