Sucrose is a non ionic compound. It does liberates ion when dissolved in water unlike NaCl or other salts which dissolve in water and produce respective cations and anions.
Thus if any amount of sucrose is dissolved in water, it will form non ionic aqueous solution (it will dissolve completely). Thus sucrose solution being non electrolytic will not conduct electricity in aqueous solution.
the bulb will not light up as sucrose will remain in molecular form only
Answer:
9
Explanation:
The structure of fluorophore used in the experiments has been drawn in the attachment. And from the drawing counting we can say that there are 9 sp2-hybridized carbon atoms present. Fiuorophores are a fluorescent chemical compound that can re-emit light upon light excitation. Normally used to produce absorbance and emission spectra.
I’m pretty sure it is A at least that’s what we did at our school to test this
Explanation:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
The atomic mass of isotope (II) of vanadium is 49.944 amu.
The question only asks regarding the direction of the equilibrium reaction. The general expression of Kp is:
Kp = [PCl₅]/[PCl₃][Cl₂]
The higher the value of K (greater than 1), the more spontaneous the reaction (favors the product side). Otherwise, it favors the reactant side. Since Kp = 0.087 which is less than 1, the direction favors the forward reaction towards the product side.
