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3 years ago

Argon (Ar): [Ne]3s23p6 core electrons: valence electrons:

Chemistry

2 answers:

Artyom0805 [142]3 years ago

6 0

Answer : Core electrons = 10

Valence electrons = 8

Explanation :

Core electrons : They are the electrons in an atom that do not participate in bonding.

Valence electrons : They are electrons in an atom that participate in bonding.

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

The electronic configuration of argon (Ar) will be:

$1s^22s^22p^63s^23p^6$

The electronic configuration of neon (Ne) will be:

$1s^22s^22p^6$

So, the electronic configuration of argon (Ar) in noble gas notation will be:

$[Ne]3s^23p^6$

In the given electronic configuration of argon, the electrons present in 1st shell and 2nd shell are the core electrons and the electrons present in 3rd shell are the valence electrons. That means,

Number of core electrons = 2 + 2 + 6 = 10

Number of valence electrons = 2 + 6 = 8

irga5000 [103]3 years ago

3 0

10 core elections

8 valence elections

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3 years ago

A piece of copper absorbs 5000 J of energy and undergoes a temperature change from 100 °C to 200 °C. What is the mass of the pie

irakobra [83]

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B.) 129.9 grams

Explanation:

To find the mass, you need to use the following equation:

Q = mcΔT

In this equation,

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-----> m = mass (g)

-----> c = specific heat (J/g°C)

-----> ΔT = change in temperature (°C)

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Q = mcΔT <----- Equation

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1 year ago

For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a

scoray [572]

Answer: The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)$

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Given: $-\frac{d[I^-]}{dt}]$ = $2.4\times 10^{-3}mol/Ls$

$+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls$

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3 years ago

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umka2103 [35]

Answer:

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Explanation:

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Step 2: Given data

We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.

The enthalpy change of combustion, given here as Δ H , tells us how much heat is either absorbed or released by the combustion of one mole of a substance.

In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number), 802.3 kJ of heat.

Step 3: calculate the enthalpy change for 3 moles

The -802 kj is the enthalpy change for 1 mole

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The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ

5 0

3 years ago

Argon (Ar): [Ne]3s23p6 core electrons:__ valence electrons:__

Argon (Ar): [Ne]3s23p6 core electrons: valence electrons: