Answer:
25.7 mL
Explanation:
Step 1: Given data
- Initial concentration (C₁): 0.350 M
- Final volume (V₂): 600 mL
- Final concentration (C₂): 0.150 M
Step 2: Calculate the volume of the initial solution
We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.150 M × 600 mL / 0.350 M
V₁ = 25.7 mL
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
Answer:
B is correct → since temperature can be measured confidently and the variations with temperature are greater ( then the errors in measuring time diminishes)
Explanation:
A is not correct → since the sugar cubes will be still surrounded by water at the temperature chosen regardless of the size of the glass, organising the data by size of the glass is not correct since the glass size does not affect in a significant way to the results ( it could affect the cooling rate of water due to the exposed surface , it but would be insignificant)
C is not correct → since the sugar cubes are approximately of the same size , putting 2 cubes or "n" cubes will not affect the time significantly since each cube is still surrounded by water at the same temperature and behaves independently from others , then each cube dissolves at the same time ( there would be small variations due to the different sizes of the cubes and the small variation due to the limited mass of water)
D is not correct → since the solubility of sugar in water is high , the amount left after the experiment would be very small ( thus the relative errors are high) and the sugar would be contaminated with water in the weighting operation, leading to more errors
B is correct → since temperature can be measured confidently and the variations with temperature are greater ( then the errors in measuring time diminishes)
Carbon: C
Oxygen: O2
C + O2 —> CO2
Reactants Product
Answer:
There are now two oxygen atoms on the left and four on the right (in one N2O and three H2O's), so we balance the oxygen atoms by placing a 2 in front of the O2.
