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2 years ago

What causes hail when it rains?

1 answer:

6 0

Simple ish explaination: the raindrops falling from a height at a fast speed whilst it is fairly cold will cause some rain to harden and freeze, becoming hail, and hail can differ in size

You might be interested in

Compare la configuración realizada con la que aparece en la tabla periódica y confirma si esta en

mart [117]

Answer:

Explanation:

4 0

2 years ago

Write a ground state electron configuration for each neutral atom

Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0

2 years ago

Is H3O+ polar or non polar? (explain please.)?

pishuonlain [190]

The overall molecule is Polar because the shape of the molecule is Trigonal Pyramidal, which means it has the lone pair electrons. Becuase of the lone pair the pulling is unequal.

H3O+ has 3 polar bonds.
To know if the bonds are polar or nonpolar find the difference of the element's electronegativity charge.
H has electronegativity charge of 2.2, and O has 3.4.
Always subtract the smaller number from the greater one.
So 3.4 - 2.2 = 1.2
If the difference is from 0-0.4 the bond is nonpolar, but if it's from 0.5-1.9 the bond is polar.
So, 1.2 is polar bond. So H3O+ has 3 polar bonds, and the overall molecule is polar too.

A simple way to know if it's polar or nonpolar is to draw the lewis dot structure, and use VSEPR.

8 0

2 years ago

1. A 10 g gold coin is heated from 25°C to 50°C (CAu is 0.13 J/g-°C). What is the H?

Oksanka [162]

The correct answer is + 32.5 J because heat is absorbed here by copper not evolved also after calculation we will find the H value is positive, lets calculate it:
H = m C Δt where:
m = 10 g
C = 0.13 J/g.°C
Δ t = final temperature - initial temperature = 50 - 25 = + 25 °C
so H = 10 x 0.13 x (+25) = + 32.5 J

4 0

3 years ago

Which isotope will spontaneously decay and emit particles with a charge of +2?

VikaD [51]

Answer: The correct option is 4.

Explanation: All the options will undergo some type of radioactive decay processes. There are 3 decay processes:

1) Alpha decay: It is a decay process in which alpha particle is released which has has a mass number of 4 and a charge of +2.

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

2) Beta-minus decay: It is a decay in which a beta particle is released. The beta particle released has a mass number of 0 and a charge of (-1).

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

3) Beta-plus decay: It is a decay process in which a positron is released. The positron released has a mass number of 0 and has a charge of +1.

$_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta$

For the given options:

Option 1: This nuclei will undergo beta-plus decay process to form $_{25}^{53}\textrm{Mn}$

$_{26}^{53}\textrm{Fe}\rightarrow _{25}^{53}\textrm{Mn}+_{+1}^0\beta$

Option 2: This nuclei will undergo beta-minus decay process to form $_{80}^{198}\textrm{Hg}$

$_{79}^{198}\textrm{Au}\rightarrow _{80}^{198}\textrm{Hg}+_{-1}^0\beta$

Option 3: This nuclei will undergo a beta minus decay process to form $_{56}^{137}\textrm{Ba}$

$_{55}^{137}\textrm{Cs}\rightarrow _{56}^{137}\textrm{Ba}+_{-1}^0\beta$

Option 4: This nuclei will undergo an alpha decay process to form $_{85}^{216}\textrm{At}$

$_{87}^{220}\textrm{Fr}\rightarrow _{85}^{216}\textrm{At}+_2^4\alpha$

Hence, the correct option is 4.

7 0

3 years ago

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