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Tpy6a [65]
3 years ago
10

Mature elephants consume an average of 600 pounds of food per day. If an elephant

Chemistry
1 answer:
MAXImum [283]3 years ago
5 0

Answer:

\large \boxed{\text{4 600 000 kg or 4600 Mg}}

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its  

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}

3. Convert days to pounds of feed

\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}

4. Convert pounds to kilograms and megagrams

\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

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Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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P_1 = initial pressure of gas = 0.974 atm

P_2 = final pressure of gas = 0.993 atm

V_1 = initial volume of gas = 27.5 mL

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T_1 = initial temperature of gas = 22.0^oC=273+22.0=295K

T_2 = final temperature of gas = 15.0^oC=273+15.0=288K

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\frac{0.974 atm\times 27.5 mL}{295K}=\frac{0.993 atm\times V_2}{288K}

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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
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Answer:

Sr(OH)₂ will be the limiting reagent.

Explanation:

First of all, you should know the following balanced chemical equation:

2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂

The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.      

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:

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  • carbonic acid: 2 moles

Now, you know the following masses of the elements:

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  • O: 16 g/mole
  • H: 1 g/mole

So the molar mass of strontium hydroxide is:

Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole

You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?

moles of strontium hydroxide=\frac{12.5 grams*1 mole}{121.62 grams}

moles of hydroxide= 0.103 moles

On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?

molesofcarbonicacid=\frac{0.150 L*3.5 moles}{1 L}

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Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?

molesofcarbonicacid=\frac{0.103 moles of strontium hydroxide*2 moles of carbonic acid}{2 moles of strontium hydroxide}

moles of carbonic acid= 0.103 moles

But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, <u><em>Sr(OH)₂ will be the limiting reagent.</em></u>

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