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horsena [70]
3 years ago
15

The solid anhydrous solid CoCl2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indica

tor to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.
Chemistry
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

CoCl_{2}.(H_{2}O)_{6} is formed and it has three unpaired electrons.

Explanation:

Blue CoCl_{2} reacts with water to produce pink colored CoCl_{2}.(H_{2}O)_{6} complex.

In this complex, Co^{2+} ion forms an octahedral complex with six H_{2}O ligands.

As H_{2}O is a weak field ligand therefore Co^{2+} ion remains in low spin state.

Hence electronic configuration of Co^{2+} ion in this octahedral geometry is t_{2g}^{5}e_{g}^{2} with total three unpaired electrons.

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Consider the following intermediate chemical equations.
vfiekz [6]

Answer:

-250.3kJ

Explanation:

Based in the reactions and using -<em>Hess's law-</em>:

(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ

(2) 4PCl₅(g) → P₄(s) + 10Cl₂  ΔH₂ = 3438kJ

The sum of (1) + (2) is:

4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ

Dividing this reaction in 4:

PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = <em>-250.3kJ</em>

8 0
4 years ago
A 400 ml sample of gas is heated from -20 c to 60
Viktor [21]
Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
t1=-20=253k
t2=60=333k
50x400/253=225xv2/333
7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml
6 0
3 years ago
Read 2 more answers
Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)
Digiron [165]

Answer: =176.6kJmol^{-1}

Explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of  C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

\Delta H= {\text {sum of bond energies of reactants}}-  {\text {sum of bond energies of products}}

\Delta H= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} -  {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}

=176.6kJmol^{-1}







7 0
4 years ago
If a frog initially contained 2 grams of carbon-14 and the half-life of carbon-14 is 5,730 years, how much carbon-14 remains in
andreyandreev [35.5K]
Half life is the time taken for a radioactive isotope to decay by half its original mass. In this case the half life of carbon-14 is 5.730 years. 
Using the formula;
New mass = original mass × (1/2)^n; where n is the number of half lives (in this case n=1 ) 
New mass = 2 g × (1/2)^1     
                  = 1 g
Therefore; the mass of carbon-14 that remains will be 1 g 
5 0
3 years ago
Use the periodic table to write the electron configuration for rubidium (Rb) in noble has notation
Alexxx [7]

Answer:

Rb: [Kr] 5s  

Step-by-step explanation:

Rb is element 37, the first element in Period 5.

It has one valence electron, so its valence electron configuration is 5s.

The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.

The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.

4 0
4 years ago
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