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4 years ago

___ is the sacrifice involved in making one decision over another

1 answer:

4 0

Opportunity cost

Opportunity cost is the benefit, profit or value that could have been received or obtained but instead given up in favor of another alternative or course of action. Every situation has several alternatives and every resource has several alternative uses as well. Thus, every decision comes with it an opportunity cost. <span>
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A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

rodikova [14]

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

6 0

3 years ago

Write a balanced chemical equation based on the following description:

kupik [55]

The balanced chemical equation for the reaction is

From the question given above, we were told that:

solid cesium reacts with solid cesium nitrite to form solid cesium oxide and nitrogen gas.

The equation for the reaction can be written as follow:

Caesium => Cs

Caesium nitrite => CsNO₂

cesium oxide => Cs₂O

nitrogen gas => N₂

Caesium + Caesium nitrite —> Caesium oxide + Nitrogen gas

The above equation can be balance as follow:

Cs + CsNO₂ —> Cs₂O + N₂

There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by writing 2 before CsNO₂ as shown below:

Cs + 2CsNO₂ —> Cs₂O + N₂

There are 2 atoms of Cs on the right side and a total 3 atoms on the left side. It can be balance by writing 6 before Cs and 4 before Cs₂O as shown below:

6Cs(s) + 2CsNO₂(s) —> 4Cs₂O(s) + N₂(g)

Now the equation is balanced

Learn more: brainly.com/question/11502387

5 0

2 years ago

Read 2 more answers

Hees anothor one theres more to

s344n2d4d5 [400]

3. all behaviors

4. hunting in packs

5 0

4 years ago

Read 2 more answers

How many liters of carbon dioxide will be produced when 89.5 L of ethane are burned? (One mole of any gas occupies 22.4 L under

yan [13]

Answer:

179 L of CO2

Explanation:

Given the equation of the reaction;

C2H6(g) + 7/2 O2(g) -------> 2CO2(g) + 3H2O(g)

Now 1 mole of ethane yields 2 moles of CO2 from the balanced reaction equation

1 mole of a gas occupies 22.4 L volume so,

22.4 L of ethane yields 44.8 L of CO2

89.5 L of ethane yields 89.5 * 44.8/22.4 = 179 L of CO2

4 0

3 years ago

What volume would 3.01•1023 molecules of oxygen gas occupy at STP?

Sliva [168]

First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2

Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2

6 0

4 years ago