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Katen [24]
3 years ago
15

1.what happens in the nucleus of an atom when an alpha particle is emitted?

Chemistry
2 answers:
Vanyuwa [196]3 years ago
8 0

Answer: 1. The mass number is reduced by 4 units and atomic number is reduced by 2 units

2. Mass number remains same but the atomic number is increased by 1.

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

_A^Z\textrm{X}\rightarrow _{A-2}^{Z-4}+_2^4\alpha

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as \beta ^--decay. In this the mass number remains same but the atomic number is increased by 1.

_A^Z\textrm{X}\rightarrow _{A+1}^Z\textrm{Y}+_{-1}^0e

coldgirl [10]3 years ago
7 0
1) The nucleus of an atom loses 2 protons and 4 neutrons.
2) The nucleus of an atom gains a proton and it's neutrons remain the same.
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Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s)→CO2(g). Whe
MA_775_DIABLO [31]

Answer:

6.82 kg

Explanation:

Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,

mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.

The sublimation enthalpy of dry ice is 571 KJ/kg.

Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.

Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.

This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.

Now, Q' =m'L' = heat lost by water = 3892.98KJ.

And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)

Therefore, m' = 6.82 kg.  

6 0
3 years ago
What energy is required to remove the remaining electron from singly ionized helium?
skelet666 [1.2K]

To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.

The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.

Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .

Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."

To know more about electron

brainly.com/question/14135172

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3 0
1 year ago
Alex cut a pizza into 8 equal slices. He removed 2 of the slices of pizza. What is the measure of the angle made by the missing
Firlakuza [10]

Answer: 90 degrees.

Explanation:

4 0
3 years ago
Sodium hydroxide, NaOH, reacts with sulfuric acid, H,SO,, in a neutralization reaction to
RSB [31]

Answer:

10.6 grams is approximately 0.10 moles. So you would need about 0.10 moles of sulfuric acid. That converts to about 9.80 grams.

Explanation:

hi girl i also wrote this in my test today so maube i hope it is correct.

mark me as brainliest if it helped you

7 0
3 years ago
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
RoseWind [281]

Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

8 0
3 years ago
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