In normal conditions, warm water does "pile up" in the" Western Pacific Ocean.
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:
The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.
Explanation:
Given conjugate base CN⁻ => weak acid => HCN => Ka =4.9 x 10⁻¹⁰
Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴
Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻