Answer:
Total Ionic equation:
H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)
Explanation:
Chemical equation:
HNO₃ + NaOH → NaNO₃ + H₂O
Balanced chemical equation:
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
Total Ionic equation:
H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)
Net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The NO₃⁻ (aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation
Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
Answer:
False
Explanation:
Molecules of the same substance are made up of the same type of atoms and look exactly alike.
Hence, if I have two molecules of methane having exactly the same atoms of carbon and hydrogen,the both are indistinguishable from each other based on appearance.
Hence all molecules of methane are exactly alike if they are composed of atoms of the same isotope of hydrogen and carbon.
Answer:
B
Explanation:
To write the electronic configuration of an element using the preceding noble gas configuration, we simply use the noble gas in the previous period of that particular element. We do not use the noble gas of that period in which the element belongs to but the one in the preceding group.
A is wrong
This is because the noble gas in the preceding period is krypton.
In fact, the electronic configuration is [Kr] 4d105s25p3
C is wrong
The noble gas in the last period here is Radon. The electronic configuration of Radium in fact is [Rn] 7s2
D is wrong
The noble gas in the last period before the period of uranium is Radon also. In fact, the electronic configuration of the element is [Rn] 5f36d17s2
B is correct, the actual electronic configuration of the element Cesium is [Xe] 6s1
Answer:
No, Stephanie is incorrect. Formation of petroleum cannot take place under the presence of oxygen.
Explanation:
Since, the petroleum is fossil product. Fossil fuel are formed under high pressure and temperature with absence of oxygen for longer period. so the way she is performing is completely incorrect. With the presence of oxygen in no way petroleum will be formed. The temperature and pressure should be in different combination for the formation of the petroleum. Along with the layers of sediments to maintain the pressure is required.
