Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Answer:
2. Na2O
Reasoning:
Na (soduim) is the only metal out of all the elements in the answer choices, and is bonded with oxygen, a nonmetal, therefore creating an ionic bond.
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
Answer:
Nuclear fusion is a kind of nuclear reaction in which two lighter nuclei combine under suitable conditions to form heavier nuclei under suitable conditions. A lot of energy is released in this process.
Oxygen and selenium both have 6 valence electrons. They're in the same group hope this helps