Answer:
0.733 mol.
Explanation:
- From the balanced equation:
<em>2Fe₂O₃ + C → Fe + 3CO₂,</em>
It is clear that 1.0 moles of Fe₂O₃ react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.
- Since Fe₂O₃ is in excess, C will be the limiting reactant.
<u><em>Using cross multiplication:</em></u>
1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.
??? mole of C produces → 2.2 moles of CO₂.
∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.
respiratory and lymphatic
Answer:
Starch is a viable indicator in the titration process because it turns deep dark blue when iodine is present in a solution. When starch is heated in water, decomposition occurs and beta-amylose is produced
No
Magnesium Chloride is MgCl2
The method I use to name ionic compounds is 'swap and drop'
Mg oxidation number is +2 and Cl oxidation number is -1
Mg^2+ Cl^-1
'swap'
Mg^1 Cl^2
'drop'
MgCl2
Answer:
The volume of the balloon will be 5.11L
Explanation:
An excersise to solve with the Ideal Gases Law
First of all, let's convert the pressure in mmHg to atm
1 atm = 760 mmHg
760 mmHg ___ 1 atm
755.4 mmHg ____ (755.4 / 760) = 0.993 atm
922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm
T° in K = 273 + °C
28.5 °C +273 = 301.5K
26.35°C + 273= 299.35K
P . V = n . R .T
First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K
(0.993atm . 6.25L) / 0.082 . 301.5 = n
0.251 moles = n
Second situation:
1.214 atm . V = 0.251 moles . 0.082 . 301.5K
V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm
V = 5.11L
