Why are hydrogen bonds important for the properties of water?

1 answer:

5 0

Water has the special type of attraction called Hydrogen bonding. The bonds between the Hydrogen and the Oxygen in each water molecule make a super dipole because the Oxygen atom is way more electronegative than the hydrogen atom. These OH bonds can then be attracted to other H2O molecules. If you have ever poured water up to the brim and there is little bit of water that is poking above the top, hydrogen bonding keeps those water molecules from spilling

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What is Mars colonization?

serious [3.7K]

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Explanation:

3 0

2 years ago

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

4 0

3 years ago

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

madreJ [45]

Answer:

$Molar \ solubility=3.12x10^{-5}M$

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$

The equilibrium expression is:

$Ksp=[Ca^{2+}][F^-]^2$

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent $x$ is computed as follows: