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natima [27]
3 years ago
7

Uranium-235 has a half-life of 713 million years. Would uranium-235 or carbon-14 be more useful for dating a fossil from Precamb

rian Time? Explain.
Chemistry
1 answer:
9966 [12]3 years ago
6 0
Answer: Uranium-235.

Radioactive isotopes are used to determine the age of antique objects, including fossils.

The half-life time of the radioactive elements is what permits the process of dating.

The half-life of C-14 is too short to be useful to date too old objects.

Precambrian time is the most antique era. C-14 hal-life is about 5730 years and Precambrian time is millions or billions of years ago. Given that the hal-life of U-235 is 704 million years it is appropiate to date the fossils from the Precambrian era.

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andrew11 [14]

hellow! how r you? 。◕‿◕。

Explanation:

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8 0
2 years ago
Perform the following
strojnjashka [21]

Answer:

4 sig figs: 20.18

Explanation:

the smallest amount of digits was 4 in the expression

7 0
3 years ago
If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?
vitfil [10]

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also  Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U  = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a  heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 =  0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1   ⇒ 69.72 + 24.2 = 93.92 °C

6 0
3 years ago
A sample of gas contains 0.1700 mol of NH3(g) and 0.2125 mol of O2(g) and occupies a volume of 17.8 L. The following reaction ta
telo118 [61]

Answer:

The volume of the sample after the reaction takes place is 19.78 L.

Explanation:

The given variables are;

Number of moles of NH₃(g) = 0.1700 mol

Number of moles of O₂(g) = 0.2125 mol

Volume occupied by the mixture = 17.8 L

The reaction

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Then takes place

That is 4 moles of NH₃(g) reacts with 5 moles of O₂(g) to produce 4 moles of NO(g) and 6 moles of H₂O(g).

Since there are less number of moles of NH₃(g) (= 0.1700 mol) in the mixture, we factor the above equation by the number of moles of NH₃(g)  present.

That is,

1 moles of NH₃(g) reacts with 5/4 moles of O₂(g) to produce 1 moles of NO(g) and 3/2 moles of H₂O(g).

Therefore,

0.1700 mol of NH₃(g) reacts with 5/4×0.1700  moles of O₂(g) to produce 0.1700  moles of NO(g) and 3/2×0.1700  moles of H₂O(g).

Which gives

0.1700 mol of NH₃(g) reacts with 0.2125  moles of O₂(g) to produce 0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Therefore, all of the NH₃(g) and O₂(g)  are consumed in the reaction and the present gases in sample then becomes

0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Total number of moles of reactant = 0.17 + 0.2125 = 0.3825

Total number of moles of product formed = 0.17 + 0.255 = 0.425

However, Avogadro's law states that equal volume of all gases at the same temperature and pressure contains equal number of molecules.

That is volume occupied by  0.3825 moles of gas = 17.8 L

Therefore the volume occupied by  0.425 moles of gas = 17.8×0.425/0.3825 L = 19.78 L

3 0
3 years ago
What is the pOH of a solution that has an H+=7.7x10^-7 M
pentagon [3]

Answer:

Explanation:

Step 1

The question is based on the concept of PH and pOH calculations.

pH is defined as negative logarithmic of hydronium Ion concentration.

while pOH is defined as negative logarithmic of hydroxide ion concentration of the solution.

Step 2

[H+] = 7.7*10-7  M  

pH = -log[H+]  

    = -log ( 7.7*10-7 )

     = 6.12

Step 3

pOH = 14 - pH

   = 14 - 6.11

  = 7.89

4 0
3 years ago
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