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Answer:
Draw circles to represent the electron shell of each atom overlapping the circles where the atoms are bonded. Add dots to represent the outer electrons of one type of atom (H). Add crosses to represent the outer electrons of the other type of atom (Cl). Make sure the electrons are always in pairs.
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:
Beryllium has an electronic configuration of 1s2 2s2 in its natural state. When in an ion state of Be2+ it loses the electrons in a 2s shell and has a configuration of 1s2. This means that Be2+ has 2 electrons.
