Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
Answer:
d
Explanation:
pv=nrt
2.5×1.01×10^5×8×10^-3=3×8.31×T
T=
Answer:
8 oxygen atoms
Explanation:
One molecule of oxygen gas has two oxygen atoms (O2). If there are four molecules, then we multiply two by four to get 8 atoms.
Answer:
E = 3.81×10 ⁻²¹ J
Explanation:
Given data:
Frequency of photon = 5.75 ×10¹² Hz
Plancks constant = 6.626 ×10⁻³⁴ Js
Energy of photon = ?
Solution:
E = h×f
E = 6.626 ×10⁻³⁴ Js × 5.75 ×10¹² s⁻¹
E = 38.1×10 ⁻²² J
E = 3.81×10 ⁻²¹ J