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
Answer:
A
Explanation:
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plant are made up of organs like roots, leaves, etc. so it can't be C.
skin (is an organ) is made up to 7 layers of ectodermal tissue so isn't B.
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Answer:
20.4 grams Zn
Explanation:
To find the mass, you first need to find the moles. This can be found using the Ideal Gas Law equation:
PV = nRT
In this equation,
-----> P = pressure (atm)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)
-----> T = temperature (K)
Before you can plug the values into the equation, you need to convert Celsius to Kelvin.
P = 0.980 atm R = 0.08206 atm*L/mol*K
V = 7.80 L T = 25.0 °C + 273.15 = 298.15 K
n = ? moles
PV = nRT
(0.980 atm)(7.80 L) = n(0.08206 atm*L/mol*K)(298.15 K)
7.644 = n(24.466)
0.312 moles = n
Now that you have the number of moles, you can convert it to grams using the atomic mass of zinc. The final answer should have 3 sig figs to match the sig figs in the given values.
Atomic Mass (Zn): 65.380 g/mol
0.312 moles Zn 65.380 grams
------------------------- x ------------------------- = 20.4 grams Zn
1 mole
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