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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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A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the
Sveta_85 [38]

Answer:

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

Explanation:

Using the formula

Ca Va = Cb Vb

Cb = 0.32 M

Vb = 50 mL = 50/1000 = 0.050L

Ca = 0.5 M

Va =?

Substituting for Va in the equation, we obtain:

Va = Cb Vb / Ca

Va = 0.32 * 0.05 / 0.5

Va = 0.016 / 0.5

Va = 0.032 L

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

6 0
3 years ago
When 250 ml of water is added to 35 ml of 0.2 M HCl
Paladinen [302]
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
8 0
3 years ago
Read 2 more answers
NO FILES
MrMuchimi

Answer:

the general equation is: A + X → AX. Where a single compound on the reactant side breaks down into two or more products during a chemical change. The general equation is AX → A + X.

Explanation:

6 0
3 years ago
According to the following reaction, how many moles of carbon dioxide will be formed upon the complete reaction of 22.7 grams of
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3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

4 0
3 years ago
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