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NeTakaya
3 years ago
9

Select the correct answer.

Chemistry
2 answers:
timofeeve [1]3 years ago
4 0

Answer:

D

Explanation:

mario62 [17]3 years ago
3 0

Answer:Answer:

I believe the wavelength is D.

Explanation:

It's the distance from wave to wave, it can be measured from peak to peak.

I hope I helped, please correct me if I'm wrong!

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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.

Know more about Book Value here:

brainly.com/question/23057744

#SPJ4

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
What is combustion in physics
NISA [10]
Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent that produces energy, usually in the form of heat and light.
5 0
3 years ago
What is the concentration of a solution if 24.0 mL of H.O is needed to be added to a 100 ml solution of 6.00 M HCI?
nikdorinn [45]
The answer to this question is 40.0M
7 0
3 years ago
Experiment #1A melting point of an old sample of Naphthalene was completed and a melting range of 77-83 oC was observed and reco
White raven [17]

Answer:

%error = 0.32%

Explanation:

Let's answer both questions, by parts.

1. Percentage error:

In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.

The expression to calculate the percentage error is the following:

%Error = absolute error / actual percentage. (1)

And the absolute error is:

Abs error = actual value - experimental value  (2)

But the experimental value is a range, so we just have to get a average of that:

Exp value = 77 + 83 / 2 = 80 °C

Now the absolute error:

Abs error = 80.26 - 80 = 0.26 °C

Finally the %error:

%error = (0.26 / 80.26) * 100

<h2>%error = 0.32%</h2>

2. Meaning of melting point range and %error

The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.

The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.

3 0
3 years ago
Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.
kumpel [21]

Answer : The limiting reactant is O_2

Explanation : Given,

Mass of C_3H_8 = 30.0 g

Mass of O_2 = 75.0 g

Molar mass of C_3H_8 = 44 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_3H_8 and O_2.

\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 1 mole of C_3H_8

So, 2.34 moles of O_2 react with \frac{2.34}{5}\times 1=0.468 moles of C_3H_8

From this we conclude that, C_3H_8 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Therefore, the limiting reactant is O_2

5 0
3 years ago
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