Answer:Therefore, there must be a direct relationship between these volumes of gases and the number of molecules they contain. Avogadro's law says that: Equal volumes of different gaseous substances, measured under the same pressure and temperature conditions, contain the same number of molecules.
Explanation:I hope it works for you.
Answer:
Equilibrium shifts to the right
Explanation:
An exothermic reaction is one in which temperature is released to the environment. Hence, if the reaction vessel housing an exothermic reaction is touched after reaction completion, we will notice that the reaction vessel e.g beaker is hot.
To consider the equilibrium response to temperature changes, we need to consider if the reaction is exothermic or endothermic. In the case of this particular question, it has been established that the reaction is exothermic.
Heat is released to the surroundings as the reactants are at a higher energy level compared to the products. Hence, increasing the temperature will favor the formation of more reactants and as such, the equilibrium position will shift to the left to pave way for the formation of more reactants. Thus , more acetylene and hydrogen would be yielded
Respuesta:
968 g Ca(OH)₂
Explicación:
Paso 1: Calcular la masa de solución
Tenemos 1500 mL de una solución cuya densidad es 1.17 g/mL, es decir, 1 mL de solución tiene una masa de 1.17 g.
1500 mL × 1.17 g/mL = 1.76 × 10³ g
Paso 2: Calcular la masa de hidróxido de calcio en 1.76 × 10³ g de solución
La solución tiene una concentración de 55% en masa de hidróxido de calcio, es decir, cada 100 gramos de solución hay 55 gramos de hidróxido de calcio.
1.76 × 10³ g Solución × 55 g Ca(OH)₂/100 g Solución = 968 g Ca(OH)₂
It would be a,b,c as the answer. hope this helps!
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
