Answer:
-24.76 kJ/g; -601.8 kJ/mol
Explanation:
There are two heat flows in this experiment.
Heat from reaction + heat absorbed by calorimeter = 0
q1 + q2 = 0
mΔH + CΔT = 0
Data:
m = 0.1375 g
C = 3024 J/°C
ΔT = 1.126 °C
Calculations:
0.1375ΔH + 3024 × 1.126 = 0
0.1375ΔH + 3405 = 0
0.1375ΔH = -3405
ΔH = -24 760 J/g = -24.76 kJ/g
ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol
<u>Answer:</u> The mass of iron (III) nitrate is 11.16 g/mol
<u>Explanation:</u>
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.3556 M
Molar mass of Iron (III) nitrate = 241.86 g/mol
Volume of solution = 129.8 mL
Putting values in above equation, we get:
Hence, the mass of iron (III) nitrate is 11.16 g/mol
Answer:
3
Explanation:
You need to remember that to measure the number of unparied electrons in an atom you need to undestard its electron configuration, and the electron configuration of phosphorus is 1s2 2s2 2p6 3s2 3p3, just the last state "3p3" have unpaired electron, and because a p state can fits 6 electrons, and here are only 3, that means that those 3 are unpaired.
Answer:
4
Explanation:
in (C2H2O2)2 there are 4 hydrogen atoms
A. <span>principal energy level, sublevel, orbital, electron</span>
