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Anvisha [2.4K]
3 years ago
12

How many protons , neutrons and electrons are in an atom of magnesium-24?

Chemistry
1 answer:
Minchanka [31]3 years ago
8 0
Here is your answer:

Theirs 12 protons, 12 electrons, and 14 neutrons!

Reason: When you look at the atomic number for any element on the table with all of the elements that's how many protons, and electrons their are in the substance (or element) you find how many neutrons by rounding the number under the atomic number which will equal 14!

Your answer is 12:12:14
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What type of compound is this?
vichka [17]

Answer:sorry id ont known

Explanation:give m brianlest if you want

6 0
3 years ago
2. The last electron in arsenic occupies a 4p level. What block
Romashka-Z-Leto [24]

Answer:

Arsenic is present in p block and group fifteen of periodic table.

Explanation:

The atomic number of Arsenic is 33.

According to the Aufbau principal in ground state of elements electron first occupy the lower energy level then fill the higher energy levels. We know that there four subshells s, p, d and f. The maximum number of electrons in these subshells can be calculated by following formula:

2 (2l +1 )

and l = 0,1,2,3,....

maximum numbers of electrons in s subshell are,

l=0

2 ( 2(0) + 1)

2

so maximum electrons in s subshell are 2.

maximum numbers of electrons in p subshell are,

l = 1

2 ( 2(1) + 1)

2( 2 + 1)

6

so maximum electrons in p subshell are 6.

maximum numbers of electrons in d subshell are,

l = 2

2 ( 2(2) + 1)

2( 4 + 1)

10

so maximum electrons in d subshell are 10.

maximum numbers of electrons in f subshell are,

l = 3

2 ( 2(3) + 1)

2( 6 + 1)

14

so maximum electrons in f subshell are 14.

Electron first fill 1s subshell then 2s subshell and in this way they goes to higher energy levels.

Electronic configuration of Arsenic:

1s2 2s2 2p6 3s2 3p6 4s2 3d10, 4p3

The last electron is present in p subshell that's way Arsenic is present in p block of periodic table.

P block elements are non-metals, metals and metalloids. These are thirty five elements. The P-block elements are present on right side of periodic table. There valance electrons are present in P orbital. The p-block metals are shiny and good conductor of heat and electricity. These metal lose the electron which is accept by non metals and form ionic bond. They have high melting points.

Metalloids includes boron, silicon, germanium, arsenic, antimony and tellurium. Metalloids contain both the properties of metals and non metals, Some metalloids are toxic like arsenic.

Most of p-block elements are non metals. They are bad conductor of heat and electricity and have low boiling points.

6 0
3 years ago
What are the two parts of a solution?
Vlad [161]
The 2 parts or components that make up a solution would be the solute and the solvent.
6 0
3 years ago
What is the correct order of energy transformation in a windmill?
Vika [28.1K]

Answer:

A wind turbine transforms the mechanical energy of wind into electrical energy. A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion.

hope it helps (^^)

# Cary on learning

8 0
2 years ago
At 25 ∘ C , the equilibrium partial pressures for the reaction 3 A ( g ) + 4 B ( g ) − ⇀ ↽ − 2 C ( g ) + 3 D ( g ) were found to
VMariaS [17]

<u>Answer:</u> The standard Gibbs free energy of the given reaction is 6.84 kJ

<u>Explanation:</u>

For the given chemical equation:

3A(g)+4B(g)\rightleftharpoons 2C(g)+3D(g)

The expression of K_p for above equation follows:

K_p=\frac{(p_C)^2\times (p_D)^3}{(p_A)^3\times (p_B)^4}

We are given:

(p_A)_{eq}=5.70atm\\(p_B)_{eq}=4.00atm\\(p_C)_{eq}=4.22atm\\(p_D)_{eq}=5.52atm

Putting values in above expression, we get:

K_p=\frac{(4.22)^2\times (5.52)^3}{(5.70)^3\times (4.00)^4}\\\\K_p=0.0632

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314 J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_{eq} = equilibrium constant at 25°C = 0.0632

Putting values in above equation, we get:

\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.0632)\\\\\Delta G^o=6841.7J=6.84kJ

Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ

6 0
4 years ago
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