The answer from this questions is the letter
B. He should not use his opinions as evidence.
Oxygen (6O2) and Glucose (C6H12O6)
<span>Reference: 6CO2 + 6H2O + light energy = C6H12O6 + 6O2.</span>
2 shells because if you do the electronic configuration:
2,7 which adds up to 9
7 stands for the group it in and it also stands for how many electrons are in the outer shell.
the amount of spaces stands for which period its in therefore it in period 2
Answer:
It is equal to the number of moles of acid that reacted. When Oxalic acid is your limiting reactant it is the # of moles of oxalic acid used. When NaOH is your limiting reactant it is equal to the number of moles of NaOH used.
Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:
PV=nRT
Solving for n,
n = PV/RT = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(25+273 K)
n = 7.897×10⁻³ mol H₂
The molar mass of H₂ is 2 g/mol.
Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>
