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svp [43]
1 year ago
13

Radioactive elements decay at a know rate known as half-life. Choose all of the correct statements concerning half-life. Carbon-

14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years.
Chemistry
2 answers:
Lera25 [3.4K]1 year ago
8 0

Answer:

-Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years.

-Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years.

Explanation:

9966 [12]1 year ago
6 0

Explanation:

Half life is simply the amount of time it takes for half of a substance to decompose.

Options;

- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.

- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000

- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)

- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)

- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)

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Vlad1618 [11]

Answer:

The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of FeNCS^{2+} as indicated on the spectrum. The reaction between the Fe^{3+} and the SCN^{-} are both clear liquids that form the orange liquid product  FeNCS^{2+} which creates the absorbance spectrum. Because the color of the solution is orange, it reflects this and similar colors while absorbing blueish hues. We can find the absorption of only the FeNCS^{2+} by pre-rinsing the cuvette with each solution we intend to measure before placing it in the spectrometer. Also, wipe each cuvette with a kimwipe to remove all fingerprints that could effect the data collection.

Explanation:

The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of FeNCS^{2+} as indicated on the spectrum.

3 0
1 year ago
An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

8 0
2 years ago
10. A solution contains 130 grams of KNO3 dissolved in 100 grams of water When 3 more grams of KNO3 is added, none of it dissolv
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Answer:

Option B

Explanation:

We will check the solubility graph for potassium nitrate,  KNO 3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.

Hence, option B is correct

3 0
2 years ago
Read 2 more answers
What is the total E associated with one mole of photons (a unit known as the Einstein) of 3.91x1019 Hz?
sashaice [31]

Answer:

        E=1.56\times 10^{10}J

Explanation:

The<em> energy of a photon</em>, E, can be calculated with the Planck-Einstein equation:

          E=hf

Where:

  • h is Planck's constant 6.626×10⁻³⁴ J.s, and
  • f is the frequency of the photon or electromagnetic radiation.

Substituting with your data:

          E=6.626\times 10^{-34}J.s\times 3.91\times 10^{19}s^{-1}=2.5908\times 10^{-14}J

Now multiply by Avogadro's number to obtain the energy of one mole of photons:

          E=2.5908\times 10^{-14}J\times 6.022\times 10^{23}=1.56\times 10^{10}J

8 0
1 year ago
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

8 0
2 years ago
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