Answer:
The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Explanation:
Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen using Lead (IV) oxide as a catalyst.
- The catalyst surface area is directly proportional to the reaction rate
- So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.
2. Also, Removing lead (IV) oxide from the reaction mixture the reaction rate decreased because as the catalyst is removed.
3. Using 50 cm³ of hydrogen peroxide doesn't affect the rate because the concentration of the reactant doesn't change.
4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased
So, The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Answer:
How do the water molecules closest to the burner compare to the water molecules closest to the surface of the liquid? ... They are closer together and moving slower than those at the surface. They are more spread out and moving faster than those at the surface. The diagram shows the sun's interior.
Explanation:
Pressure<span> with Height: </span>pressure<span> decreases with increasing </span>altitude<span>. The </span>pressure<span>at any level in the atmosphere may be interpreted as the total weight of the </span>air<span>above a unit area at any </span>elevation<span>. At higher elevations, there are fewer </span>air<span>molecules above a given surface than a similar surface at lower levels</span>
The mass of 
that would be formed will be 18.22 grams
<h3>Stoichiometric calculations</h3>
Let us first look at the balanced equation of the reaction:
The mole ratio of Y to 
is 2:3.
Mole of 10.0 grams of Y = 10/88.9 = 0.11 moles
Mole of 10.0 grams = 10/71 = 0.14 moles
3/2 of 0.11 = 0.165. Thus, is limiting in availability.
Mole ratio of and = 3:2
Equivalent mole of = 2/3 x 0.14 = 0.093 moles.
Mass of 0.093 moles =0.093 x 195.26 = 18.22 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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Explanation:
Son muchos los beneficios del catión monovalente (K+) en las plantas y cultivos, mencionaremos algunos:
-Interviene en el proceso de apertura y cierre de los estomas; el catión (k+) ayuda a que los estomas puedan realizar correctamente el proceso de la fotosíntesis es decir la absorción del dióxido de carbono para producir oxígeno y agua; este proceso ayuda a que la planta pueda fabricar su propio alimento es decir transformar materiales inorgánicos en orgánicos como azúcares y carbohidratos que son aprovechables por la planta para realizar diversas funciones catalíticas como la respiración celular,la fosforilación y la síntesis de proteínas; este proceso también contribuye a que la planta pueda almacenar y retener suficiente agua lo cual ayuda a evitar el estrés hídrico en la planta.
Por lo tanto si la planta tiene una buena concentración de potasio como catión monovalente(k+); es posible que tenga un buen crecimiento y desarrollo radicular ,así mismo la calidad de los cultivos será muy buena,seran cultivos muy nutritivos y saludables. Algunos cultivos que necesitan una alta cantidad de potasio son los cítricos y el banano.
