1 molecule of glucose contains 6 atoms of C, 12 atoms of H , and 6 atoms of 0.1 mole of glucose contains 6 moles of C atoms , 12 moles of H atoms , and 6 moles of O atoms .
Answer:Score” scatter plot shows an example of a positive relationship—as one variable increases, so does the other. The points in this type of scatter plot tend to go “uphill” from left to right
Explanation:Score” scatter plot shows an example of a positive relationship—as one variable increases, so does the other. The points in this type of scatter plot tend to go “uphill” from left to right
Answer:
There will not be any ejection of photoelectrons
Explanation:
Energy of the photon= hc/λ
Where;
h= Plank's constant
c= speed of light
λ= wavelength of the incident photon
E= 6.6×10^-34 × 3 ×10^8/488 × 10^-9
E= 4.1 ×10^-19 J
Work function of the metal (Wo)= 2.9 eV × 1.6 × 10^-19 = 4.64 × 10^-19 J
There can only be ejected photoelectrons when E>Wo but in this case, E<Wo hence there will not be any ejection of photoelectrons.
Answer:
C₆H₈O₆
Explanation:
The molecular formula of given compound is C₆H₈O₆. Because there are six carbon, six oxygen and eight hydrogen atom present in this compound.
The molecular weight of following compound is,
C₆H₈O₆
C = 12 amu
H = 1 amu
O = 16 amu
C₆H₈O₆ = 12×6 + 1×8 + 16×6
C₆H₈O₆ = 72 +8 + 96
C₆H₈O₆ = 176 g/mol
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