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Marysya12 [62]
3 years ago
13

For a price ceiling to be a binding constraint on the market, the government must set it a above the equilibrium price. b below

the equilibrium price. c precisely at the equilibrium price. d at any price because all price ceilings are binding constraints.
Chemistry
1 answer:
beks73 [17]3 years ago
4 0

Answer:

B. Below the equilibrium price

Explanation:Price ceiling is the legal price,impose by The Government or Regulatory agencies above which a product should not be sold. It is also a price control mechanism through which the Government and other Regulatory agencies check the excesses of producers and the middle men(Whole sellers and retailers). The price ceiling has to be fixed below the equilibrium price for it to be binding, because businesses are interested mainly for profit maximization.

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One gram of salt in 100 liters of water could be considered a _______________________ solution. A) concentrated B) dilute C) sat
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Answer:I believe your answer would be option B) dilute. Hope this helps.

Explanation:

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3 years ago
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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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Answer:

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Answer:

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What is the volume of a rock that has a mass of 50 grams and a density of<br> 5 g/ml?
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