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maria [59]
2 years ago
4

A piece of charcoal used for cooking is found at the remains of an ancient campsite. a 0.94 kg sample of carbon from the wood ha

s an activity of 1580 decays per minute. find the age of the charcoal. living material has an activity of 15 decays/minute per gram of carbon present and the half-life of 14c is 5730 y. answer
Chemistry
1 answer:
Tems11 [23]2 years ago
3 0
Mass of sample of charcoal = 0.94 kg = 0.00094 

∴, <span>activity = decay rate / mass = 1580/0.00094
                                                 = 1.681 X 10^6 decays per min per gram


Using the half-life formula, we have: 
activity of sample / activity of modern carbon = (1/2)^(age / half-life) 
</span>∴, Age = <span>half-life x log (base 2) (modern activity / coal activity)</span>
           = 5730 x log(base 2)(1.681X10^6/ 15)<span>
           = 96115 years.

</span><span>Answer: Age of the charcoal = </span>96115 years
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Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0
2 years ago
Which nuclear emission moving through an electric field would be deflected toward the positive electrode?
Tom [10]
As a rule, opposite charges attract. So,you would expect that the particle that would deflect towards the positive electrode has a negative charge. Now, an alpha particle is basically a neutral Helium atom. Beta particle is an electron. Gamma particle has no charge (neutral). <em>Thus, the answer is B.</em>
6 0
3 years ago
A chemist mixes 75.0 g of an unknown substance at 96.5C w/ 1,150 g of water at 25.0C . if the final temperature of the system is
vichka [17]
Remember: heat lost = heat gained 

When calculating heat loss or gain, remember 

mass*(spec heat cap)*(change in T) 

The unknown loses heat- we don't know the spec heat cap, so we'll call it x.

The water gains. I've omitted the units, but always use when solving problems on your own. 

75*x*(96.5-37.1) = 1150*4.184*(37.1-25) 
<span>
Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
3 0
2 years ago
A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg
Dafna1 [17]

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

8 0
2 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
VLD [36.1K]

Answer:- 64015 J

Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.

density of water is 1 g per mL.

So, the mass of water = 4250mL(\frac{1g}{1mL})  = 4250 g

Final temperature of water after adding the hot copper bar to it is 26.15 degree C.

So, \Delta T for water = 26.15 - 22.55 = 3.60 degree C

Specific heat for water is 4.184 \frac{J}{g.^0C}

The heat gained by water is calculated by using the formula:

q=mc\Delta T

where, q is the heat energy, m is mass and c is specific heat.

Let's plug in the values in the formula and do the calculations:

q=4250g*\frac{4.184J}{g.^0C}*3.60^0C

q = 64015 J

So, 64015 J of heat is gained by the water.



5 0
2 years ago
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