In radiation therapy,

The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that mac

KonstantinChe [14]

The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine: TRUE

<h3>What is the book value?</h3>

Book value is the worth of an asset based on its balance sheet account balance in accounting.
The value of an asset is determined by subtracting the asset's original cost from any depreciation, amortization, or impairment expenses.
Traditionally, a company's book value is equal to its total assets minus intangible assets and liabilities.
In practice, however, depending on the source of the computation, book value may include either goodwill or intangible assets, or both.
The value inherent in its employees, which is part of a company's intellectual capital, is always overlooked.
When intangible assets and goodwill are specifically omitted, the indicator is frequently defined as "tangible book value."

Therefore, the statement "the book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine" is TRUE.

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Complete question:

The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine. (Ignore taxes.) TRUE or FALSE

7 0

2 years ago

Any help would be appreciated. Confused.

masya89 [10]

Answer:

q(problem 1) = 25,050 joules; q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ... Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) = m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) = m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve. Good luck on your efforts. Doc :-)

5 0

3 years ago

using the following thermochemical equation, determine the amount of heat produced from the combustion of 24.3 g benzene (c6h6).

aliya0001 [1]

the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ

There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.

Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).

Thus, 24.3 g of benzene contains:

n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.

2 moles = 6278 kJ

0.311 moles =x

By the straightforward direct three rule:

2x = -1953.08 x = -976.5 kJ

Learn more about combustion here-

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4 0

1 year ago

I really need help with this Im lost

Illusion [34]

I believe it would weigh less if it’s blown up with helium because helium is lighter than air

5 0

3 years ago

The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28

ivanzaharov [21]

<u>Answer:</u> The mass of second isotope of indium is 114.904 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

.....(1)

Let the mass of isotope 2 of indium be 'x'

<u>For isotope 1:</u>

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

<u>For isotope 2:</u>

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

$114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu$

Hence, the mass of second isotope of indium is 114.904 amu

4 0

2 years ago