Answer:
There is 117.4 kJ of heat absorbed
Explanation:
<u>Step 1: </u>Data given
Number of moles CS2 = 1 mol
Temperature = 25° = 273 +25 = 298 Kelvin
Heat absorbed = 89.7 kJ
It takes 27.7 kJ to vaporize 1 mol of the liquid
<u>Step 2:</u> Calculate the heat that is absorbed
C(s) + 2S(s) → CS2(l) ΔH = 89.7 kJ (positive since heat is absorbed)
CS2(l) → CS2(g) ΔH = 27.7 kJ (positive since heat is absorbed)
We should balance the equations, before summing, but since they are already balanced, we don't have to change anything.
C(s) + 2S(s)---> CS2 (g)
ΔH = 89.7 + 27.7 = 117.4 kJ
There is 117.4 kJ of heat absorbed
This question does not contain the structures of the molecules. The structures in Daylight SMILES format are:
I. C1=CC=CC=C1C(=O)C
II. C1=CC=CC=C1CC=O
III. C1=CC(C)=CC=C1C=O
IV. C1=CC=CC=C1CCC
V. C1=CC=CC=C1C(C)C
The structures are also attached
Answer:
The structure of compound IV is consistent with the information obtained analysis
Proposed structures for the ions with m/z values of 120, 105,77 and 43 are (also attached):
C1=CC=CC=C1C(=[OH0+])C |^1:7|
C1C([CH0+]=O)=CC=CC=1
C1[CH0+]=CC=CC=1
C(#[OH0+])C
respectively
Explanation:
The IR peak at 1687 cm⁻¹ is indicative of an α unsaturated carbonyl carbon. While the 1H NMR singlet is of the methyl group next to carbonyl and the multiplet near 7.1 ppm is a characteristic peak of benzene. This data shows points towards structure I.
Mass spectrum peak at 120 m/z is of molecular ion peak. In the case of carbonyl-containing molecule, this peak is observable. The signal at 105 shows the loss of a methyl group next to the carbonyl. m/z value of 77 is the characteristic cationic peak of benzene, while the peak at 43 infers the formation of acylium ion (RCO+) due to α-cleavage. All this data agrees with the structure of acetophenone (Structure 1)
Applications of iron oxide nanoparticles include terabit magnetic storage devices, catalysis, sensors, superparamagnetic relaxometry (SPMR), and high-sensitivity biomolecular magnetic resonance imaging (MRI) for medical diagnosis and therapeutics.
Answer is: 9623.85 kJ of heat is <span>transferred from iron ingot.
</span>m(Fe) = 24.7 kg · 1000 g/kg = 24700 g; mass of iron ingot.
C = 0.4494 J/g°C; t<span>he specific heat of iron
</span>ΔT = 880°C - 13°C; temperature <span>difference.</span>
ΔT = 867°C.
Q = m·C·ΔT.
Q = 24700 g · 0.4494 J/g°C ·867°C.
Q = 9623856.06 J ÷ 1000J/kJ.
Q = 9623.85 kJ.
Answer:
24.9mL of the stock solution are required
Explanation:
To solve this question we have to find, as first, the moles of HCl that we need to prepare the desire solution. These moles are taken from the stock solution as follows:
<em>Moles diluted solution:</em>
500.0mL = 0.5000L * (0.613mol / L) = 0.3065 moles HCl
As these moles comes from the 12.3M HCl solution, the volume that we need of the stock solution is:
<em>Volume stock solution:</em>
0.3065 moles HCl * (1L / 12.3moles) = 0.0249L 12.3M HCl =
<h3>24.9mL of the stock solution are required</h3>
