In the periodic table the lanthanoid and the actinides are place separately at the bottom because of their electronic configuration and their properties compared to the other elements.
The the lanthanoid and the actinides are place separately at the bottom in the periodic table due to their electronic configuration and the properties. and to make the periodic table more convenient . if we place the f block elements that is he lanthanoid and the actinides then the size of the periodic table will increase. the f block elements are called as the inner transition element.
Thus , to make the periodic table more convenient and to group the elements in the block the the lanthanoid and the actinides are place separately at the bottom.
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Answer:
I screenshot and expained here
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Explanation:
There are 17.97 moles of calcium chloride would react with 5. 99 moles of aluminum oxide .
The balanced chemical equation between reaction between calcium chloride and aluminum oxide is given as,
→ 
The molar ratio of above reaction is 3:1
It means 3 moles of calcium chloride is require to react one mole of aluminum oxide.
The number of moles of calcium chloride requires to react with 5. 99 moles of aluminum oxide = 3 × 5. 99 = 17.97 moles
The equation in which number of atoms of elements in reactant side is equal to the number of atoms of elements in product side is called balanced chemical equation .
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Explanation:When the stove turns on the skillet will become very hot because of the conduction of heat transferring from the stove to the skillet.
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
