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NikAS [45]
2 years ago
7

An atom has an average atomic mass of about 24.3 amu. What is the chemical

Chemistry
2 answers:
Genrish500 [490]2 years ago
6 0

Answer: Option (a) is the correct answer.

Explanation:

Atomic number is the sum of only total number of protons present in an element. Whereas mass number is the sum of total number of both protons and neutrons present in an element.

So, for an atom with atomic mass 24.3 amu must have an atomic number as 12.

And, according to the periodic table only magnesium is the element which has an atomic number 12. As every element has different atomic number.

Chemical symbol which represents magnesium is Mg.

Thus, we can  conclude that Mg is the chemical  symbol for the atom which has an average atomic mass of about 24.3 amu.

USPshnik [31]2 years ago
3 0
So first off, Look at the periodic table and find which of the elements has the amu of 24.3, amu is just the number below the element and the answer would be Mg
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A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me
DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0
2 years ago
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In a first order reaction 40% of reactant gets converted into product in 30 minutes. What time would it require to convert 75% i
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56 minutes and 15 seconds I believe
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A student dissolved a sample in hexane, spotted it on to a TLC plate and eluted using ethyl acetate. After visualizing the TLC p
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1) Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:

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2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?

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3) After a rather lengthy organic chemistry synthesis procedure, a student ran the product of the reaction on a TLC plate and obtained the result below. What might he/she have done wrong, if anything?

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4) A student spots an unknown sample on a TLC plate. After developing in hexanes/ethyl acetate 50:50, he/she saw a single spot with an Rf of 0.55. Does this indicate that the unknown material is a pure compound? What can be done to verify the purity of the sample?

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6) Plate A, below, represents the TLC chromatogram of a compound run in hexanes. The same compound was then spotted on a large TLC plate and again run in hexanes. Which TLC plate, B, C, or D, correctly represents how far the compound would run on the longer plate?

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Back to TLC

Original content © University of Colorado at Boulder, Department of Chemistry and Biochemistry.
The information on these pages is available for academic use without restriction.
3 0
2 years ago
A sample taken from a layer of mica in a canyon has 2.10 grams of potassium-40. A test reveals it to be 2.6 billion years old. H
DanielleElmas [232]

Answer:

\boxed{ \text{8.40 g}} 

Explanation:

The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

 No. of                                 Fraction

<u>half-lives</u>      <u>    t/yr      </u>       <u>Remaining</u>  

      0               0                          1

      1                1.3  billion              ½

      2              2.6                          ¼

      3              3.9                          ⅛

We see that after 2 half-lives, ¼ of the original mass remains.

Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.

Original  mass = 4 × 2.10 g = \boxed{ \text{8.40 g}}

7 0
2 years ago
Acetaldehyde shows two UV bands, one with a lmax of 289 nm ( 5 12) and one with a lmax of 182 nm ( 5 10,000). Which is the n p*
UkoKoshka [18]

Answer:

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

Explanation:

The two types of acetaldehyde transition are as follows:

n→π* and π→π*

From the attached diagram we have to:

ΔEn→π* < ΔEπ→π*

ΔEα(1/λ)

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λn→π* > λπ→π*

In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.

The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

5 0
2 years ago
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