Answer:
3.37 × 10²³ molecules
Explanation:
Given data:
Mass of C₆H₁₂O₆ = 100 g
Number of molecules = ?
Solution:
Number of moles of C₆H₁₂O₆:
Number of moles = mass/molar mass
Number of moles = 100 g/ 180.16 g/mol
Number of moles = 0.56 mol
Number of molecules:
1 mole contain 6.022 × 10²³ molecules
0.56 mol × 6.022 × 10²³ molecules /1 mol
3.37 × 10²³ molecules
Since the direction of particle displacement in electromagnetic waves is also perpendicular to the direction of motion, generating the waveform of visible light and other forms of electromagnetic radiation, they are also transverse waves.
In a transverse wave, the displacement is perpendicular to the direction of motion (at an angle of 90 degrees Celsius). The direction of displacement (up and down) in the case of the ocean wave is perpendicular to the direction of wave motion (horizontally along the water), making it a transverse wave.
How far a particle has moved from its original starting position, or, in the case of an ocean wave, how high or low the water is, is measured by its displacement or amplitude.
learn more about displacement here;
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Answer:
Sugar is a food/spice. Macromolecules are the tiniest molecules if i was right. I do'nt know whether it is right. Happy to help...
Explanation:
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
D. or C. I think it is more D. than anything else
