<span>If a molecule wants to interact with an electric field. it should have a permanent dipole momentum. so first check the polarity. for example CH4 is not polar. CH2Cl2 is polar (so yes), H3O)+ is obvious that is polar. H2O yes. C2H2Cl2 in trans form is not polar but in cis form yes. CO2 is non-polar. Ozone is polar. </span>
- The change in color from blue to pink of the cobalt complexes here has been the basis of cobalt chloride indicator papers for the detection of the presence of water. It is also used in self-indicating silica gel desiccant granules.
- Pink cobalt species + chloride ions ⇌ Blue cobalt species + water molecules
<u>Explanation</u>:
- The adjustment in color from blue to the pink of the cobalt complexes here has been the premise of cobalt chloride indicator papers for the detection of the presence of water. It is likewise utilized in self-demonstrating silica gel desiccant granules.
Pink cobalt species + chloride particles ⇌ Blue cobalt species + water molecules
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The response of [Co(H2O)6]2+(aq) + 4Cl–(aq) → [CoCl4]2–(aq) + 6H2O(l) is endothermic. In this manner, as per Le Chatelier's rule, when the temperature is raised, the situation of the balance will move to one side, shaping a greater amount of the blue complex particle at the expense of the pink species.
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Including concentrated hydrochloric raises the chloride particle fixation, making the equilibrium move to one side, as per Le Chatelier. Including water brings down the chloride particle fixation, moving the equilibrium the other way.
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As an extension, it is conceivable to show that it is the Cl–particles in the hydrochloric acid that move the balance by including a spatula of sodium chloride rather than the pink arrangement. This delivers a bluer color, however, this may take some time because the salt is delayed to dissolve.
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
.</span>
Answer:
1.33 atm
Explanation:
use general gas equation P1 V1/ T1 = P2 V2/ T2
rearrange and make P2 the subject then solve,it should give you 1.33 atm
Answer:
2 H2(g) + O2(g) → 2 H2O(ℓ) ΔH = −570 kJ
Explanation:
