Answer:
Keq = 0.053
7.3 kJ/mol
Explanation:
Let's consider the following isomerization reaction.
glucose 6‑phosphate ⇄ glucose 1 - phosphate
The concentrations at equilibrium are:
[G6P] = 0.19 M
[G1P] = 0.01 M
The concentration equilibrium constant (Keq) is:
Keq = [G1P] / [G6P]
Keq = 0.01 / 0.19
Keq = 0.053
We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.
ΔG° = -R × T × lnKeq
ΔG° = -8.314 J/mol.K × 298 K × ln0.053
ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol
The answer would be A.Bias because the scientist can form a Bias opinion based on his beliefs
2200 mg of antibiotic
Explanation:
Given that 40 mg of antibiotic/kg of the bodyweight is given.
If patient is 55 kg then the dose of antibiotic will be
if 40/1000000 is done then we can get antibiotic in kg/kg of the weight
= 0.00004 kg of antibiotic per kg
0.00004*55 ( to know how much 55 kg person will require)
= 0.0022 kg
This 0.0022 value will be converted to mg
0.0022*10^6
= 2200 mg of antibiotic will be given to a 55kg patient.
Answer:
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Explanation:
