atoms consist of protons and neutrons in the nucleus , surrounded by electrons that reside in orbitals . Orbitals are classified according to the four quantum number s that represent anyone particular orbitals energy , shape , orientation and the spin of the occupying electron.
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Answer:
Mole fraction = 0,0166
Explanation:
Mole fraction is defined as mole of a compound per total moles of the mixture. In the solution, the solute is fructose and the solvent is water. That means you need to find moles of fructose and moles of water.
The molecular mass of fructose is 180,16g/mol and mass of water is 18,02 g/mol. Using these values:
91,7g fructose × (1mol / 180,16g) = <em>0,509 moles of fructose</em>
545g water × (1mol / 18,02g) = <em>30,24 moles of water</em>
Thus, mole fraction of fructose is:
<em>Mole fraction = 0,0166</em>
I hope it helps!
Animal cells have a centrosome and lysosomes, whereas plant cells do not.
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
The generalized rate expression may be written as:
r = k[A]ᵃ[B]ᵇ
We may determine the order with respect to B by observing the change in rate when the concentration of B is changed. This can be done by comparing the first two runs of the experiment, where the concentration of A is constant but the concentration of B is doubled. Upon doubling the concentration of B, we see that the rate also doubles. Therefore, the order with respect to concentration of B is 1.
The same can be done to determine the concentration with respect to A. The rate increases 4 times between the second and third trial in which the concentration of B is constant, but that of A is doubled. We find that the order with respect to is 2. The rate expression is:
r = k[A]²[B]
