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mario62 [17]
3 years ago
5

Ethylenediamine is a bidentate ligand. The oxalate group, used in this experiment was also a didentate ligand. The structure of

ethylenediamine is NH2CH2CH2NH2. The oxalate group formed a coordinated compound using the negative charges on the oxygen. Explain how the ethylenediamine compound will bond.
Chemistry
1 answer:
Murljashka [212]3 years ago
5 0

Answer:

Ethylene diamine will bond to the Central metal via a lone pairs of electrons on nitrogen

Explanation:

Complexes are formed by coordinate bond formation. Before a coordinate bond is formed, one of the species must have a lone lair of electrons available for donation into empty orbitals on the central metal.

Ethylene diammine contains nitrogen which has a lone pair of electrons. The two lone pairs on the two nitrogen atoms can bond with the central metal. This makes ethylene diammine a bidentate ligand (two bonding atoms).

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Electron configurations are a shorthand form of an orbital diagram, describing which orbitals are occupied for a given element.
kipiarov [429]

Answer:

[Ar] 3d10 4s2 4p3 is shorthand, 1s22s22p63s23p63d104s24p3 long hang

Explanation:

The shorthand is made using the lowest & closest noble gas, and picking up where it leaves off as follows, and longhand is made from a followed pattern you can easily find

3 0
3 years ago
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A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
DerKrebs [107]

Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
5 0
3 years ago
A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled
dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$     \left(\frac{\text{mass}}{\text{volume}}\right)

Now after adding extra 10 mL , the density becomes $\frac{m}{60}$.

Therefore, $\frac{m}{50}>\frac{m}{60}$

So the density decreases when we add more solution.

4 0
3 years ago
Isotopes are different types of atoms of the same element, but with a different number of __________ ?
rodikova [14]
Same things but with different numbers of neutrons in nuclei
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3 years ago
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The specific heat of gold is 0.031 calories/gram°C and the specific heat of silver is 0.057 calories/gram°C. If equal amounts of
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I think the substance that will heat up faster would be the silver metal since it has a higher heat capacity. Heat capacity is the amount of heat needed to raise the temperature of the system into one degree. Heat capacity and heat energy is directly related so higher value of heat capacity would lead to higher heat energy.
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3 years ago
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