Answer: 1.48 atmosphere
Explanation:
Pressure in kilopascal = 150
Pressure in atmosphere = ?
Recall that 1 atmosphere = 101.325 kilopascal
Hence, 1 atm = 101.325 kPa
Z atm = 150 kPa
To get the value of Z, cross multiply
150 kPa x 1 atm = 101.325 kPa x Z
150 kPa•atm = 101.325 kPa•Z
Divide both sides by 101.325 kPa
150 kPa•atm/101.325 kPa = 101.325 kPa•Z/101.325 kPa
1.48 atm = Z
Thus, 150 kPa is equivalent to 1.48 atmospheres
The smaller number is the number of protons, and the greater number is the mass.
Answer:
The answer is hardness and brittleness. The ion discharging procedure and taking after impact on reservoir water quality file, for example, hardness, pH, conductivity, has been dissected. Comes about uncovered that the most discharge quality of various materials was touching base at the 30s after startup.
Explanation:
Answer: It will take 8.2 minutes until the concentration decreases to 0.055 M
Explanation:
The time after which 99.9% reactions gets completed is 40 minutes
Explanation:
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.


b) Time taken for 0.085 M to decrease to 0.055 M


Thus it will take 8.2 minutes until the concentration decreases to 0.055 M
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:
![= \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)

0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J