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Ann [662]
3 years ago
14

What may be expected when K < 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is

small. The concentration of one or more of the products is small. The reaction will not proceed very far to the right. The reaction will generally form more reactants than products.
Chemistry
1 answer:
mestny [16]3 years ago
4 0

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

  • We have fewer products than reactants
  • We have more reactants than products
  • The position of equilibrium lies to the left

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

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Molar mass is used in a conversion factor to calculate ___ and uses the units of ____.
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Answer:

The molar mass is used in a conversion factor to calculate the weight of one mole of a given compound and uses the units of grams / mole ____.

Explanation:

Se utilizan los pesos atómicos (Z) de cada componente para calcular la masa de 1 mol de compusto, ejemplo:

H20 (agua)

Masa molar H20 = (Peso atomico H) x 2 + Peso atomico 0 = 1, 007g x2 + 16g = 18 gramos / mol

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You are provided with a compound fertilizer, 40-15-10. Calculate the quantity of fertilizer to add to a one hectare field to sup
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Answer:

The correct answer is a) 300 Kg b) 225 Kg c) 400 Kg and d) 600 Kg.

Explanation:

Based on the given value, that is. 40-15-10 shows the compositions of nitrogen, phosphorus, and potassium found in the compound fertilizer in the form of percentage.  

41 percent nitrogen is equal to 40/100 = 0.4, 15 percent phosphorus is equal to 15/100 = 0.15, and 10 percent potassium is equal to 10/100 = 0.1.  

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a) At 120 Kg per ha, the required nitrogen will be 120/0.4 = 300 Kilogram of fertilizer.  

b) At 90 kg per ha, the required nitrogen will be 90/0.4 = 225 Kilograms of fertilizer.  

c) At 60 kg per ha, the required phosphorus will be 60/0.15 = 400 Kilograms of fertilizer.  

d) At 60 Kg per ha, the required potassium will be 60/0.1 = 600 Kilograms of fertilizer.  

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