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Black_prince [1.1K]
3 years ago
14

The product of the nuclear reaction in which 40Ar is subjected to neutron capture followed by alpha emission is ________. The pr

oduct of the nuclear reaction in which 40Ar is subjected to neutron capture followed by alpha emission is ________. 37S 45Ca 36S 35Ar 41Ar
Chemistry
1 answer:
Orlov [11]3 years ago
5 0

Answer:

37S

Explanation:

Radioactivity is the spontaneous emission of particles and / or electromagnetic radiation by unstable atomic nuclei leading to their disintegration.

We have two main types of radioactivity: radioactive decay and artificial transmutation.

In radioactive decay ( natural radioactivity ), a naturally occurring radioactive element like Uranium-238 disintegrates or decays into more stable isotopes with the emission of particles and/or radiation.

23892U = 23490Th + 42He

Artificial transmutation is the collision of two particles where one particle captures the other used to bombard it. There is subsequent production of isotopes similar or different from the bombarded particle. Neutrons, alpha particles ( helium nucleus ), electrons, protons can be used to bombard elements.

147N + 42He = 178O + 11P

For the above question which is artificial transmutation, the reaction equation is

4018Ar + 10n = 3716S + 42He

So, the neutron capture by Argon-40 will produce a radioisotope Sulphur-37 with the emission of an alpha particle.

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I am not for sure, but i think it explodes :)
3 0
3 years ago
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4Cr(s)+3O2 (g )= Cr2O3 (s)<br> Find the theoretical yield in moles and grams
Assoli18 [71]
The balanced reaction is:

<span>4Cr(s)+3O2 (g )= Cr2O3 (s)

Since we are not given the amount of any of the reactants, we assume we have one gram of chromium. Calculations are as follows:

1 g Cr ( 1 mol Cr / 52 g Cr ) ( 1 mol Cr2O3 / 4 mol Cr ) = <span>0.005 mol Cr2O3

</span></span>0.005 mol Cr2O3 (151.99 g Cr2O3 / 1 mol Cr2O3 ) = 0.7307 g <span>Cr2O3
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Therefore, the theoretical yield for 1 gram of Cr is 0.005 mol Cr2O3 or 0.7307 g Cr2O3.
7 0
3 years ago
It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2
Lapatulllka [165]

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

n_{HCl}=2*n_{Mg(OH)_2}

And in terms of volumes and concentrations we verify:

V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}

So we use the given data to proof it:

4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

Best regards.

4 0
3 years ago
What are two examples of an endothermic reaction and what are two examples of an ectothermic reaction?
natka813 [3]
Exothermic processes: Making ice cubes,formation of snow in clouds


Endothermic processe: Melting ice cubes, evaporation of water
7 0
4 years ago
A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point
Ksivusya [100]

Answer:

P_{T} = 2.94 atm

Explanation:

The total pressure (P_{T}) in the container is given by:

P_{T} = P_{O_{2}} + P_{He}

The pressure of the oxygen (P_{O_{2}}) and the pressure of the helium (P_{He}) can be calculated using the ideal gas law:

PV = nRT

<u>Where</u>:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

n_{O_{2}} = \frac{m}{M}

Where m is the mass of the gas and M is the molar mass

n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles  

And the number of moles of helium is:

n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles

Now, we can find the pressure of the oxygen and the pressure of the helium:

P_{O_{2}} = \frac{nRT}{V} = \frac{1.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 0.98 atm

P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm

Finally, the total pressure in the container is:

P_{T} = P_{O_{2}} + P_{He} = 0.98 atm + 1.96 atm = 2.94 atm

Therefore, the total pressure in the container is 2.94 atm.

I hope it helps you!

6 0
4 years ago
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