Answer:
6⅔ shifts
Explanation:
From the question given:
A shift = 4 hours
Pay = $8.25 per hour
Next, we shall determine the number of hours that will result in a pay of $220. This can be obtained as follow:
$8.25 = 1 hour
Therefore,
$220 = $220 × 1 hour / $8.25
$220 = 220/8.25 hours.
$220 = 80/3 hours
$220 = 26⅔ hours
Therefore, it will take 26⅔ hours to receive a pay of $220.
Finally, we shall determine the number of shifts in 26⅔ hours. This can be obtained as follow:
4 hours = 1 shift
Therefore,
26⅔ hours = 26⅔ ÷ 4
26⅔ hours = 80/3 × 1/4
26⅔ hours = 80/12
26⅔ hours = 20/3
26⅔ hours = 6⅔ shifts
Therefore, she will work 6⅔ shifts in order to receive a pay of $220
Vegetable soup should be the correct answer. The reason for this is that it can easily be split, where as the rest would be very difficault if not imposible because of chemical change.
I believe the correct answer would be that b<span>oiling points and melting points are similar because they both involve the change in a state of a material, but they are different because boiling point involves a change from a liquid to a gas and melting point involves a change from a solid to a liquid. Boiling and melting are phase changes that can happen to a substance however they differ in the process that happens.</span>
The chemical formula is
composed of hydrogen and oxygen where the two atoms are bonded through hydrogen
bond type of bonding. In this pair, the oxygen is the more electronegative atom
hence the electrons are more directed to it. Because of this, this creates
electron polarity which affects the chemical property of water.
<span> </span>
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:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)
= 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
