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3 years ago

Which compound matches the ir spectrum best?

1 answer:

4 0

Small peak at 3000large peak at 1685F: it contains two benzene rings that is connected by a bunch of carbons and ketone-Explanation: The spectrum shows a stretching absorption consistent with a ketone functional group: carbonyl C=O stretching at ~1685 cm-1. (An aldehyde, by contrast, would also show a ~2700 cm-1 absorption for the carbonyl C-H stretch.) The C=O stretching frequency is consistent with an aromatic ketone, such as in compound F (1,4-diphenyl-1,4-butanedione). In contrast, an aliphatic ketone absorbs at higher energy (~1710 cm-1). The spectrum also shows the typical ~1600 & ~1500 cm-1 absorptions of a phenyl group.

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Which orbital is partially filled in the Zirconium (ZI) atom?

Nikolay [14]

Answer:

4d orbital.

Explanation:

Hello!

In this case, since zirconium's atomic number is 40, we fill in the electron configuration up to 40 as shown below:

$1s^2,2s^2,2p^6, 3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^2$

Thus, the orbital 4d is partially filled.

Best regards!

3 0

3 years ago

An object has a mass of 441 g and a volume

Ratling [72]

Answer:

<h3>The answer is 4.41 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{441}{10} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

---------------------------------------------------------------------------------

Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$

Plug in all the given values in the above equation -

$E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V$

So, $E_{cell}=2.32V$

3 0

3 years ago

Calculate the volume of a 3.50 M solution of H2SO4 made from 49 g of H2SO4

maxonik [38]

Answer:

0.143L

Explanation:

Molar mass of H2SO4 = 98g/Mol

No of mole = mass/molar mass

No of mole= 49/98 = 0.5 mol

No of mol = concentration × volume

Volume = n/C = 0.5/3.5 = 0.143L

3 0

3 years ago

What are the roles of producers, primary consumers, secondary consumers, tertiary consumers for the trophic levels in the energy

Nesterboy [21]

This question is more for Biology than Chemistry, but the role of producers is to make energy (food) to be consumed. In a pyramid diagram, the producers would be at the bottom. Now going up the pyramid, the primary conumers are the first to consume producers and obtain energy from them. As you go up the pyramid, the secondary consumers will consume the primary consumers as a way to obtain energy, and the same goes for tertiary consumers towards secondaries.

As you go up the energy pyramid, you will notice a trend that there is less energy being obtained from each consumer. In other words, the producers will ALWAYS have more energy than the tertiary consumers.

I hope this answers your question.

3 0

3 years ago