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katen-ka-za [31]
3 years ago
10

In any engineering design problem, the first step is to understand the problem and identify one or more possible solutions. In t

his task, you’ll analyze the problems you face as the chemical engineer challenged with setting up the ammonia-making process. Recall the chemical equation for producing ammonia:
N2 + 3H2 ⇌ 2NH3 + energy

1.Explain the problem surrounding the ammonia-making process in terms of chemical equilibrium.
Chemistry
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

Explanation:

About 2,070 results (0.60 seconds)  

Answer is: at lower temperatures the reaction rate would decrease. The lower is the temperature, the slower the reaction becomes. ... Because this is exothermic reaction (enthalpy is less than zero), at lower temperatures, the equilibrium is in favor of ammonia, but the reaction doesn't proceed at a detectable rate.

Hi,

let us go through the question again

N2 + 3H2 ⇌ 2NH3 + energy  

Explain the problem surrounding the ammonia-making process in terms of chemical equilibrium.

So I would approach this question on the basis of effect of each of the reactants and products concentration on the equilibrium synthesis of ammonia.

If the concentration of any reactant is increased, the yield of ammonia is increased.

If the concentration of ammonia is reduced by removing it as it forms, the yield will as well be increased.

This reaction also yields heat meaning it is an exothermic reaction, so when we increase the temperature, the yield will be reduced. Hence it is favored by low temperature.

You might be interested in
What determines how fast a substance will dissolve
Blababa [14]

Answer:

(1) the surface area of the solute,

(2) the temperature of the solvent,

(3) the amount of agitation that occurs when the solute and the solvent are mixed.

Explanation:

8 0
4 years ago
Read 2 more answers
When a 2.00 g sample of KCl is dissolved in water in a calorimeter that has a total heat capacity of 1.28 kJ ⋅ K − 1 , the tempe
Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

q=c\times (\Delta T)

where,

q = heat produced = ?

c = specific heat capacity of water = 1.28kJ/K

\Delta T = change in temperature = 0.360 K

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

q=1.28kJ/K\times 0.360K

q=0.4608kJ=460.8J

Now we have to calculate the molar heat solution of KCl.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 460.8 J

m = mass of KCl = 2.00 g

Molar mass of KCl = 74.55 g/mol

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole

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

\Delta H=\frac{460.8J}{0.0268mole}

\Delta H=17194.029J/mol=17.19kJ/mol

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

7 0
3 years ago
Carbordum is silicon carbide SiC a very hard material used as an abrasive on sand paper and in other applications. it is prepare
AleksandrR [38]

Answer:

4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.

Explanation:

The balanced equation for the reaction between silicon dioxide and carbon at high temperature is given as:

SiO_2+3C\rightarrow SiC+2CO

1 mole silicon dioxide reacts with 3 moles of carbon to give 1 moles of silicon carbide and 2 moles of carbon monoxide.

Mass of SiC = 3.00kg = 3000.00 g

1 kg = 1000 g

Molecular mass of SiC = 40 g/mol

Moles of SiC = \frac{3000.00 g}{40 g/mol}= 75 mol

According to reaction, 1 mole of SiC is produced from 1 mole of silicon dioxide.

Then 75 moles of SiC will be produce from:

\frac{1}{1}\times 75 mol=75 mol of silicon dioxide.

mass of 75 moles of silicon dioxde:

75 mol\times 60 g/mol=4500 g=4.5 kg

4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.

3 0
4 years ago
Please help with 3! Please give only the correct answer...
cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
_________________________________________________________
Explanation: 
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
______________________________________________________
We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
__________________________________________________
Then we write:  "  * 10^ "
__________________________________________________
   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
      "pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).   
______________________________________________________
Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
             →  "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
 "NEGATIVE TEN" { "-10" } .

So we write this value as:  " 1.75 * 10^(-10)  m " .  

{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
______________________________________________________
The answer is:  " 1.75 * 10^(-10)   m " .
______________________________________________________
4 0
4 years ago
Which action do researchers take to make advances in science?
melamori03 [73]
<span>The action that researchers take to make advances in science would be conducting experiments to test their hypothesis. By doing such, they are able to know whether the hypothesis is true or not. Hope this answers the question. Have a nice day.</span>
3 0
3 years ago
Read 2 more answers
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