Hello!
At
Standard Pressure and Temperature, an ideal gas has a molar density of
0,04464 mol/L.So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.
So, the Density of Sulfur Dioxide (SO₂) at STP is
2,8599 g/LHave a nice day!
Answer:
Option b. 0.048 M
Explanation:
We have the molecular weight and the mass, from sulcralfate.
Let's convert the mass in g, to moles
1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.
Molarity is mol /L
Let's convert the volume of solution in L
10 mL . 1L/1000 mL = 0.01 L
4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L
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
Answer:
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