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maria [59]
3 years ago
7

Organize the following atoms according to increasing number of orbitals surrounding the nucleus: Rubidium (Rb), Phosphorous (P),

Neon (Ne), Gallium (Ga)

Chemistry
1 answer:
enot [183]3 years ago
5 0
Answer is: Ne, P, Ga, Rb.
₁₀Ne 1s²2s²2p⁶. 
₁₅P 1s²2s²2p⁶3s²3p³.
₃₁Ga 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p¹.
₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹.
Neon has smallest number of orbitals because neon has smallest atomic number (ten electrons), rubidium has largest number of orbitals because rubidium has largest atomic number.
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1. What property do elements share if they are on the same row of the periodic table? What if they are in the same column?
Nimfa-mama [501]
If they are in the same row they have the same number of outer electrons on the same main shell also known as valence electrons

if they are in the same column they have the same number of valence electrons but on different main shells
8 0
3 years ago
What do all fossils fuels have in common
lys-0071 [83]

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7 0
2 years ago
How many kJ of heat are released by the reaction of 25.0 g of Na2O2(s) in the following reaction? (M = 78.0 g/mol for Na2O2)
solong [7]

-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.

Explanation:

Given:

mass of Na2O2 = 25 grams

atomic mass of Na2O2 = 78 gram/mole

number of mole = \frac{mass}{atomic mass of 1 mole}

                          = \frac{25}{78}

                          =0. 32 moles

The balanced equation for the reaction:

2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ

It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.

similarly 0.3 moles of Na2O2 on reaction would give:

\frac{126}{2} = \frac{x}{0.32}

x = \frac{126 x 0.32}{2}

 = -20.16 KJ

Thus, - 20.16 KJ of energy will be released.

6 0
3 years ago
Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that
Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻     E° = 2.38 V       oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s)      E° = -0.25 V     reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)      E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0
3 years ago
A sample of 0.600 mol of a metal m reacts completely with excess fluorine to form 46.8 g of mf2. how many moles of f are in the
oksian1 [2.3K]
The balanced chemical reaction is expressed as:

M + F2 = MF2

To determine the moles of the element fluorine present in the product, we need to determine the moles of the product formed from the reaction and relate this value to the ratio of the elements in MF2. We do as follows:

moles MF2 produced = 0.600 mol M ( 1 mol MF2 / 1 mol M ) = 0.600 mol MF2
molar mass MF2 = 46.8 g MF2 / 0.6 mol MF2 = 78 g/mol
moles MF2 = 46.8 g ( 1 mol / 78 g ) = 0.6 mol
moles F = 0.6 mol MF2 ( 2 mol F / 1 mol MF2 ) = 1.2 moles F
6 0
3 years ago
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