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Katena32 [7]
3 years ago
10

Which complex ion can exhibit geometric isomerism? Assume that M is the metal ion, A and B are ligands, and the geometry is octa

hedral. a) [MA.12+ b) [MAB]2+ c) [MA4B2]2+ d) [MA3B312+
Chemistry
1 answer:
Archy [21]3 years ago
7 0

Answer:

[MA4B2]2+

Explanation:

Octahedral species are species that has a coordination number of of six. This means that six ligands are attached to the central metal atom/ion.

For any octahedral specie to exhibit geometrical or cis-trans isomerism, it must have the formula EX2Y4 as in [MA4B2]2+, hence the answer above.

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What is the answer if u know please answer
forsale [732]
An atom having 52 protons and 54 electrons would have an atomic number of 52 and a net charge of -2. This element would be 52 Te 2-, or choice A.
7 0
4 years ago
Describe some of the dangers of Global warming.
grin007 [14]

Answer:

Increased Heat, Drought, and Insect outbreaks.

Explanation:

Increased heat, drought and insect outbreaks, all linked to climate change, have increased wildfires. Declining water supplies, reduced agricultural yields, health impacts in cities due to heat, and flooding and erosion in coastal areas are additional concerns.

8 0
3 years ago
Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
3 years ago
Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide
Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

       According to question

                                2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\

Since 'NO' is the limiting reagent according to this ratio.

According to equation

         2 moles NO reacts to form 2 moles NO₂

So,  0.1855 moles NO give  = 0.1855 moles of NO₂

            Mass of 1 mole NO₂ = 46 g/mole

            Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0
3 years ago
Heliocentric theory:
Juliette [100K]
All of th above support the heliocentric theory.
7 0
3 years ago
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