Explanation:
Compounds having same molecular formula but different structural and spatial arrangement are isomers.
Three isomers are possible for dibromomethene.
In one structure (IUPAC name: 1,1-dibromomethene), both the bromine atoms are attached to one carbon atom.
In another two structures (Cis and trans), two bromine atoms are attached to two different carbon atoms.
In Cis 1,2-dibromomethene, two bromine atoms are present on the same side.
Whereas in Cis 1,2-dibromomethene, two bromine atoms are present on the opposite side and hence, does not have net dipole moment.
I’m pretty sure it’s B since ones charge the neutral while ones is positive, it’s an obvious difference as well/
Answer:
It is basic and has a pH of 9.8.
Explanation:
pOH = 4.2
we will determine its pH.
pOH + pH = 14
pH = 14 - pOH
pH = 14 - 4.2
pH = 9.8
According to pH scale the the pH lower than 7 is consider acidic while pH of seven is neutral and pH greater than 7 is basic.
The given solution has pH 9.8 it means it is basic.
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:
0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
Answer:
MOLARITY= 0.3092mol/l
ABSOLUTE UNCERTAINTY= 0.000873
Explanation:
The equation of reaction is
2HNO3 + Na2CO3 ⟶ 2NaNO3 + H2O + CO2.
QUESTION1: CALCULATION FOR MOLARITY;
Molarity= gram mole of solute ÷ liters of solution
Where;
Mole of solute= mass ÷ molar mass
Therefore;
Mole of solute= 0.8311g ÷ 105.988g/mol= 0.0078515mol
MOLARITY= 0.0078415mol ÷ 25.36ml = 0.0003092mol/ml = 0.3092mol/l
This is the Molarity of the solution
QUESTION2: CALCULATION FOR ABSOLUTE UNCERTAINTY;
Uncertainty (u) =√([0.05 ÷ 25.36]^2 + [0.001 ÷ 105.988]^2 + [0.0007 ÷ 0.8311]^2) × Molarity
Solving brackets gives
(0.00197161+0.00000943503+0.00084226) ×Molarity
Adding up gives
0.002823×Molarity
Therefore;
ABSOLUTE UNCERTAINTY= 0.002823×0.3092= 0.000873
