In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an archaeological site has a C-14 disintegration rate of 9.16 atoms per minute per gram of carbon. Estimate the age of this sample in years. The half-life of C-14 is 5730 years. (enter only the number of years in standard notation, not the unit years)

Answers

Answer 1

Answer:

Rate of disintegration is defined as the time required by a sample or substance at which half of the radioactive substance disintegrates. It depends on the nature of disintegration and amount of substance.

The age of the sample is approximately 4241.17 years.

Given that:

C-14 atoms disintegration rates = 15.3 atom/ min-g

Rate of disintegration of the sample = 9.16 atom/ min-g

The digit proportion of carbon-14 is = $(9.16)/(15.3)$ = 0.5987

Now, also the half-life of carbon-14 is 5730 years.

Such that:

$(1)/(2)^n = \text A$

$(1)/(2) = 0.5987$

Taking log:

n log 2 = -log 0.5987

Thus, n = $(0.227)/(0.3010)$

n = 0.740

The age of the sample can be given by:

Age = n x half-life

Age = 0.740 x 5730

Age = 4241.17 years.

Therefore, the age of the substance is 4241.17 years.

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Answer 2

Answer:

Answer:

The answer is "4,241 .17 years"

Explanation:

The disintegration rate, which shows in C-14 atoms = $15.3 (atoms)/(min-g)$

Rate of sample disintegration = $9.16 \frac{atoms} {min-gram}$

The digit proportion of C-14 can be determined that is included in the sample $= \frac {9.16}{15.3} \n\n = 0.5987$

5730 years from half-life.

The number with half-lives (n) which are repelled must be determined:

$((1)/(2))^n= A\n\nA= fraction of C-14, which is remaining \n\n((1)/(2))^n= 0.5987 \n\n n \log 2 = - \log 0.5987\n\n$

$\therefore \n\n \Rightarrow n= (0.227)/(0.3010) \n\n = 0.740\n$

So, the age of the sample is given by = $n *\ half-life$

$= 0.740 * 5730 \ years \n\n=4241.17 \ years\n\n$

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G write the symbols for the cation and anion that make up each ionic compound Co(NO3)2

Answers

Answer:

$Co(NO_3)_2$ is formed of cation, $Co^(2+)$ and anion, $NO_3^(-)$

Explanation:

Naming of the ionic compounds:-

The name of the cation is written first and the the name of the anion is written after the name of the cation separated by single space.
The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
In case of transition metals, the oxidation state are written in roman numerals in bracket in front of positive ions.

Hence, given ionic compound:-

$Co(NO_3)_2$ is formed of cation, $Co^(2+)$ and anion, $NO_3^(-)$

Thus, the name must be Cobalt(II) nitrate.

You would like to make a 100 mL buffer solution at pH 8.00. Assuming you would like to accomplish this with a hypochlorous acid (HOCl) buffer (HOCl/NaOCl), Ka= 3.0 * 10-8. If the solution is 0.3 M in HOCl, what concentration of NaOCl would be necessary in the buffer solution to obtain a pH of 8.0?

Answers

Answer:

To obtain the pH of 8.0, the concentration of NaOCl needs to be 0.9 M in the 0.3 M HOCl solution

Explanation:

This problem can be solved by Henderson-Hasselbalch equation, which gives relation between the concentration of acid, its salt, pKa and the pH of the solution. This equation is given as,

$pH=-log(K_a)+log([NaOCl])/([HOCl])$

By placing the known variables in the above equation we get,

$8=-log(3*10^(-8))+log([NaOCl])/(0.3)$

$8-7.52=log([NaOCl])/([0.3])$

$10^(0.48)=([NaOCl])/(0.3)$

$[NaOCl]=10^(0.48)*{0.3}$

$[NaOCl]=0.9 M$

The above calculations show that the required concentration of NaOCl is 0.9 M.

A car is traveling at 87.0 km/hr. How many meters will it travel in 37.0 seconds?

Answers

Speed of the car = $87.0 (km)/(hr)$

Given time = 37.0 s

Converting time from seconds to hours:

1 hr= 60 min; 1 min = 60 s

$37.0 s * (1 min)/(60s)*(1 hr)/(60min) = 0.0103 hr$

Calculating distance from speed and time:

$0.0103 hr * (87.0 km)/(1 hr) =0.894 km$

Converting distance from km to m:

$0.894 km * (1000 m)/(1 km) = 894 m$

So the distance traveled by the car in 37.0 s is 894 m.

In which orbitals would the valence electrons for carbon (C) be placed? s orbital and d orbitals s orbital only p orbitals only s orbital and p orbitals

Answers

Answer:

p orbitals only

Explanation:

Carbon has an atomic number of 6 so its electron configuration will be 1s² 2s² 2p². It has two orbitals as indicated with the 2 as its period number with the outer orbital have 4 valence electrons. So carbon is in the p-orbital, period 2 and in group 4.

Final answer:

Carbon's valence electrons reside in the 2s and 2p orbitals. These orbitals hybridize during bond formation to create equivalent sp3 hybrid orbitals, as evidenced in the methane molecule. Carbon's valence electrons are not placed in d orbitals.

Explanation:

Carbon (atomic number 6) has a total of six electrons. Two of these fill the 1s orbital. The next two fill the 2s orbital, and the final two are in the 2p subshell. According to Hund's rule, the most stable configuration for an atom is one with the maximum number of unpaired electrons. Therefore, carbon has two electrons in the 2s subshell and two unpaired electrons in two separate 2p orbitals. When discussing valence electrons, the electrons in the outermost shell are the ones considered, which for carbon are the electrons in the second shell namely 2s and 2p.

The geometry of the methane molecule (CH4) illustrates that in the bonding process, the s and p orbitals hybridize to allow the formation of four equivalent bonds with hydrogen atoms. Without hybridization, we would expect three bonds at right angles (from the p orbitals) and one at a different angle (from the s orbital). Nonetheless, through orbital hybridization, all four bonds in methane are identical, which is explained by the concept of sp3 hybridized orbitals.

Therefore, the valence electrons for carbon would be placed in the s orbital and p orbitals, not in the d orbitals, because carbon does not have electrons in the d subshell in its ground state. Additionally, the s and p orbitals are the only ones involved in bonding for carbon in most of its compounds, such as methane.

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Calculate the molar mass of each compound given below. c4h4

Answers

When asked to find molar mass, we first need to calculate the total number of atoms of each element in the compound.

In $C_(4) H_(4)$ , we see that there are 4 Carbons and 4 Hydrogens.

We then need to look up the atomic mass of each of these elements, which is found on the periodic table.

For carbon, the atomic mass is 12.01g
For hydrogen, the atomic mass is 1.008g

Then we multiply the number of atoms in the element by the atomic mass of the element:

C: $4*12.01g=48.04g$
H: $4*1.008g=4.032g$

And then we need to add these two values together to get the molar mass of the compound:

$48.04g+4.032g=52.072g$

So now we know that the molar mass of $C_(4) H_(4)$ is 52.072g.

Three resonance structures of the given anion are possible. One is given, but it is incomplete. Complete the given structure by adding nonbonding electrons and formal charges. Draw the two remaining resonance structures (in any order), including nonbonding electrons and formal charges. Omit curved arrows.

Answers

Answer:

Explanation:

The missing incomplete resonance structure is attached in the image below. From there, we can see the addition of the nonbonding electrons and its' formal charge which makes the resonance structure a complete resonance structure. The others two resonance structure that can be derived from the complete structure is also shown in the image. Out of these three structures, the structure that contributes most to the hybrid is the structure with the negative charge on the oxygen.

Final answer:

To complete the provided resonance structure, add nonbonding electrons and formal charges. Then, draw the two remaining resonance structures by distributing the nonbonding electrons and formal charges differently.

Explanation:

When completing the provided incomplete structure of the anion, you need to add nonbonding electrons and formal charges to make it accurate. Then, draw the two remaining resonance structures by distributing the nonbonding electrons and formal charges differently. To illustrate, let's consider the example of a nitrate ion (NO3-). The complete structure of the provided resonance form would have a double bond between the central nitrogen atom and one of the oxygen atoms, with two lone pairs on the nitrogen atom. The remaining two resonance structures would have different double bond oxygen-nitrogen combinations.

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