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Naily [24]
3 years ago
15

What do the elements in each pair have in common? k, kr be, mg ni, tc b, ge al, pb?

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0
The pairs are:
K, Kr - Same period
Be, Mg - Same group
Ni, Tc - Both are transition metals
B, Ge - Both are metaloids
Al, Pb - Both form inert oxides
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Describe how p and l wave move after an earthquake
Genrish500 [490]

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3 years ago
Benzene is an organic chemical compound with the molecular formula C6H6. In the benzene molecule, carbon atoms form a ring with
Sergeu [11.5K]

Answer:

sp^2

Explanation:

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The six-carbon benzene ring contains two types of bonds: C-C and C-H bonds, that are  sp^2-hybridized σ bonds, and the six π bonds that form the aromatic ring. The σ bonds form from one s orbital and two p orbitals from each carbon, which then bond the carbon to the two carbons on either side and the carbon's single hydrogen. The remaining p orbital from each carbon atom sticks out above and below the plane of the ring; these p orbitals overlap sideways, rather than lengthwise, to form the aromatic π bond system.

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8 0
3 years ago
Think about chemicals you may have in your home and list as many as you can think of:
MAVERICK [17]

Sucrose; C12H22O11

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6 0
3 years ago
Read 2 more answers
An interplanetary probe returns to Earth with soil samples. The atoms in the sample are sorted by their number of protons. A new
Leya [2.2K]

Answer is: the average atomic mass is 232.

ω₁ = 20% ÷ 100%.

ω₁ = 0.20.

ω₂ = 80% ÷ 100%.

ω₂ = 0.80.

Ar₁ = 120 (number of protons) + 120 (number of neutrons).

Ar₁ = 240.

Ar₂ = 120 + 110 .

Ar₂ = 230.

Average atomic mass of atoms of bolognium =  

Ar₁ · ω₁ + Ar₂ · ω₂.  

Average atomic mass of atoms of bolognium =  240 · 0.2 + 230 · 0.8.  

Average atomic mass of atoms of bolognium = 48 + 184.  

Average atomic mass of atoms of bolognium = 232.

4 0
3 years ago
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

4 0
3 years ago
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