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3 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 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

2 years ago

How much oxygen is typically required to complete a combustion reaction

DIA [1.3K]

Answer:

Mass of oxygen required = 19.5 moles × 16 g/mol = 312 g.

Explanation:

8 0

1 year ago

Write a balanced chemical equation for the following reaction in an acidic solution: Copper metal reacts with aqueous nitric aci

Andreyy89

Answer:

Cu(s) + 4HNO₃(aq) -> Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

4 0

2 years ago

If you have 20.0 g of CO2, how many atoms of Oxygen (O) are present in the sample?

Alecsey [184]

Answer:

20 g of CO₂ have 5.47x10²³ oxygen atoms

Explanation:

1 mol of CO₂ has 1 mol of C and 2 moles of O.

Mass / Molar mass = Moles

20 g / 44 g/m = 0.454 moles

In 1 mol we find 2 moles of O

In 0.454 moles of CO₂, we will find ( 0.454 .2) = 0.909 moles

1 mol has _____ 6.02x10²³ atoms

0.909 moles have (0.909 . 6.02x10²³) = 5.47x10²³ atoms

7 0

2 years ago

What is the mole fraction of cation in 1M of aluminium sulphate

Svetllana [295]

The mole fraction of cation (Al)=0.4

<h3>Further explanation</h3>

Given

1 M of Aluminium sulphate

Required

The mole fraction of cation

Solution

Ionization of the Aluminum sulfate solution( assume 1 L solution ) :

mol Al₂(SO₄)₃ = M x V = 1 M x 1 L = 1 mol

Al₂(SO₄)₃⇒2Al³⁺ + 3SO₄²⁻

1 mol 2 mol 3 mol

From this equation, total mol in solution = 2+3 = 5 moles

Mol fraction Al(as a cation) :

= 2/5=0.4

8 0

3 years ago