Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.
A chemical change
Explanation:
In this reaction, a chemical change has occurred, here are some indicators of chemical changes from the reaction;
Chemical changes are changes in which new products are formed.
- Formation of bubbles and precipitates: these are new substances that indicates a chemical change. The bubbles and insoluble precipitate are good indicators of chemical change.
- A temperature change that is large; This is also typifies chemical changes. When temperature rises in reaction, a chemical change has occurred.
- The bubbles and precipitates cannot be combined to give the reactants. The reaction is irreversible.
learn more:
Chemical change brainly.com/question/9388643
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<u>Answer:</u> The specific heat of metal is 0.821 J/g°C
<u>Explanation:</u>
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of metal = 30 g
= mass of water = 100 g
= final temperature = 25°C
= initial temperature of metal = 110°C
= initial temperature of water = 20.0°C
= specific heat of metal = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]](https://tex.z-dn.net/?f=30%5Ctimes%20c_1%5Ctimes%20%2825-110%29%3D-%5B100%5Ctimes%204.186%5Ctimes%20%2825-20%29%5D)
Hence, the specific heat of metal is 0.821 J/g°C
Only the temperature of gas
Answer:
54 g
Explanation:
1 mole of water = H2O
mass of 1 mole of H2O= mass of h2 + mass of o
= 2× mass of h +mass of o
= 2×1+16 =18 g
1 mole of water = 18g
3moles of water = 18×3g= 54g
