Somewhat false
observations can be made of a model of the statue of liberty, say, or in real line
Answer:
HgO (empirical formula)
Explanation:
4.08 - 3.78 = 0.3g (oxygen)
0.02 : 0.02
0.02/0.02 : 0.02/0.02
1 : 1 (ratio)
HgO ( empirical formula)
2HgO ----> 2Hg + O2 ( your equation correct)
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
Altitude is height above sea level. The density of air decreases with height. There are two reasons, at higher altitudes there is less air pushing down from above,and gravity is weaker farther from earths center.
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
