Answer:
The change in T° is 47.1 °C
Explanation:
Calorimetry formula to solve this:
Q = m . C . ΔT
We replace the data gien:
635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT
In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:
635 J = 35 g . 0.385 J/g°C . ΔT
635 J / 35 g . 0.385 J/g°C = ΔT
47.1°C = ΔT
El gas se puede medir por volumen en cm3 utilizando un instrumento conocido como eudiómetro.
Answer:
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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.
1. HCl (H:1, Cl 35.5) ---> 1+35.5 = 36.5
2. FeS (Fe:56, S:32) ---> 56+32 = 88
3. Cl2 (Cl:35.5) ---> 35.5 x 2 = 71
4. CaC03 (Ca:40, C: 12, O: 16) ---> 40 + 12 + 3(16) = 100
5. Fe0 (Fe: 56, O: 16) ---> 56+16=72