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ale4655 [162]
2 years ago
8

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio

of the mass of 14 6c to the total mass of carbon in the atmosphere is r, and the decay constant of 14 6c is λ. assume that, at any time, 14 6c is a negligible fraction of the total mass of carbon and that the measured activity of the sample is purely due to 14 6c. also assume that the ratio of mass of 14 6c to total carbon mass in the atmosphere (the source of the carbon in the sample) is the same at present and on the day when the number of 14 6c atoms in the sample was set. express your answer in terms of the mass ma of a 14 6c atom, mc, a, r, and λ.
Chemistry
1 answer:
Paha777 [63]2 years ago
6 0
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
 The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) = 
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

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Determine the compound type for the following formulas: C12H22011 Mg(OH)2 H20 Cu3Zn2 Au <br>​
Scorpion4ik [409]

Answer:

C12H22O11  

✔ covalent

Mg(OH)2    

✔ Ionic

H2O    

✔ covalent

Cu3Zn2    

✔ metallic

Au      

✔ metallic

Explanation:

7 0
2 years ago
Read 2 more answers
NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol×\frac{1mol}{80,043g}X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0
2 years ago
A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.145 g of this subunit w
Hitman42 [59]

Answer:

The molar mass of the protein is 12982.8 g/mol.

Explanation:

The osmptic pressure is given by:

π=MRT

Where,

M: is molarity of the solution

R: the ideal gas constant (0.0821 L·atm/mol·K)

T: the temperature in kelvins

Hence, we look for molarity:

0.138 atm=M(0.0821\frac{l*atm}{mol*K} )(28+273K)

M=\frac{0.138atm}{(0.0821\frac{l*atm}{mol*K} )(301K)}= =5.584×10⁻³mol/l

As we have 2 ml of solution, we can get the moles quantity:

Moles of protein: 5.584×10⁻³\frac{mol}{l}\frac{1l}{1000ml}×2ml=1.117×10⁻⁵mol

Finally, the moles quantity is the division between the mass of the protein and the molar mass of the protein, so:

Moles=Mass/Molar mass

Molar mass= Mass/Moles=\frac{0.145g}{1.117*10^{-5}mol}=12982.8 g/mol

8 0
2 years ago
A student mixed together aqueous solutions of Y and Z. A white precipitate(solid)formed. which could not be Y and Z
maks197457 [2]
<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)

</span><span>(B)hydrochloric acid + sodium hydroxide 
</span><span>HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)

<span>(D)sodium chloride + silver nitrate
</span>NaCl(aq) +  AgNO3(aq) ----> AgCl(s) +NaNO32(aq)

AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.

4 0
2 years ago
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the a
geniusboy [140]

Answer:

0.1M solution of NaOH

Explanation:

1 mole of NaOH - 40g

? moles - 1 g = 1/40 = 0.025 moles.

Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume

= 0.025/0.25

= 0.1M.

8 0
2 years ago
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