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Maurinko [17]
3 years ago
12

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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Answer:

There are three atoms in the equation

6 0
3 years ago
Before measuring the absorbance of a solution with the ocean optics spectrophotometer, it is important to run a blank sample of
Strike441 [17]
I think the most appropriate answer is: the solvent being used in the experiment
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5 0
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]

Answer:

B.) 129.9 grams

Explanation:

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

Q = mcΔT

In this equation,

-----> Q = energy (J)

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

-----> c = specific heat (J/g°C)

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

The specific heat of copper is 0.385 J/g°C. Knowing this, you can plug the given values into the equation and simplify to isolate "m".

Q = mcΔT                                                            <----- Equation

5000 J = m(0.385 J/g°C)(200 °C - 100 °C)        <----- Insert values

5000 J = m(0.385 J/g°C)(100)                            <----- Subtract

5000 J = m(38.5)                                                <----- Multiply 0.385 and 100

129.9 = m                                                             <----- Divide both sides by 38.5

8 0
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)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}

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

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

6 0
3 years ago
1. Methane (CH4) combusts as shown in the equation below. If 3 moles of methane were used in the combustion, what would be the c
umka2103 [35]

Answer:

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

Explanation:

<u>Step 1: </u>The balanced equation

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<u>Step 2:</u> 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 <u>one mole</u> 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.

<u>Step 3: </u>calculate the enthalpy change  for 3 moles

The -802 kj is the enthalpy change for 1 mole

The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ

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

5 0
3 years ago
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