I have attached an image of the IR spectrum required to answer this question.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.
Correct Answer: option <span>(1) Mn(s)
Reason:
The </span><span>spontaneity of electrochemical cell, depends on the it's Eo value. Electrochemical cells with positve Eo are spontanous and vice-versa.
</span>
In present case, the Eo of half-cell of interest are as follows:
Eo Zn2+/Zn = <span>-0.763v
</span>Eo Mg2+/Mg = 2.37v
Eo Mn2+/Mn = -1.18v
Therefore, Eo cell (with Zn as one of the half-cell) = Eo Zn2+/Zn - Eo Mn2+/Mn
= -0.763 - (-1.18)
= 0.417v
On other hand, Eo cell (with Mg as one of the half-cell) = Eo Mg2+/Zn - Eo Mn2+/Mn
= -2.37 - (-1.18)
= -1.19v
Thus, Mn(s) <span>metal will spontaneously react with Zn2+(aq), but will not spontaneously react with Mg2+(aq)</span>
Answer:
The valence electrons in the metal atom in the lattice structure can delocalize and move freely. The electrons are "loosely" connected with their parent atom in the metallic bond because most metals have excess electrons in their outer orbitals above the stable configuration.
Explanation:
Says google
Chemistry is the only type of science that studies how substances change.
Hope this helps you :)