Answer:
B. wave 1 has a larger wavelength than wave 2
<u>Answer:</u> The freezing point of solution is -0.454°C
<u>Explanation:</u>
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
The equation used to calculate depression in freezing point follows:
To calculate the depression in freezing point, we use the equation:
Or,
where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 2
= molal freezing point elevation constant = 1.86°C/m
= Given mass of solute (KCl) = 5.0 g
= Molar mass of solute (KCl) = 74.55 g/mol
= Mass of solvent (water) = 550.0 g
Putting values in above equation, we get:
Hence, the freezing point of solution is -0.454°C
Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Explanation:
B. a circle graph
circle graphs are the best to show percentages because they’re very easy to look at and get info from
The number of moles of NH3 that could be made would be 0.5 moles
<h3>Stoichiometric reactions</h3>
From the balanced equation of the reaction:
N2 (g) + 3 H2(g) ----> 2NH3 (g)
The mole ratio of N2 to H2 is 1:3
Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.
Mole ratio of H2 and NH3 = 3:2
Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:
2 x 0.75/3 = 0.5 moles
More on stoichiometric calculations can be found here: brainly.com/question/8062886