Considering change of state descriptions, select the name of the change of state for when energy is absorbed. Group of answer ch
1 answer:
You might be interested in
solution:
The question is incomplete, here is the complete question:
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide:
In the second step, acetylene, carbon dioxide and water react to form acrylic acid:
Write the net chemical equation for the production of acrylic acid from calcium carbide, water and carbon dioxide. Be sure your equation is balanced.
<u>Answer:</u> The net chemical equation is written below.
<u>Explanation:</u>
The intermediate balanced chemical reaction are:
(1) ( × 6 )
(2)
To omit acetylene from the net chemical reaction, we multiply Equation (1) by 6.
<u>Equation 1:</u>
<u>Equation 2:</u>
<u>Net chemical equation:</u>
Hence, the net chemical equation is written above.
Answer : The question is to write the elements alphabetically and seperating them with commas.
So in this case for your question of Fireworks , K
,
The answer will be K, N, O.
You have to just differentiate the elements and write them according to the alphabetical order.
So in this case for your question of Fireworks , K
The answer will be K, N, O.
You have to just differentiate the elements and write them according to the alphabetical order.
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant can be calculated as:
= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant can be calculated as:
0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J