The ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.
<h3>What is a NMR spectrum?</h3>
Nuclear magnetic resonance spectroscopy is a spectroscopy that shows the detailed structure and chemical environment of a chemical element.
Pentan-3-ol contain 12 hydrogen atoms. In H-NMR spectra, hydrogen atoms have same environment gives a signal.
There are 4 different kinds of signals due of the 4 different environment experienced by these 12 hydrogens.
Thus, the ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.
Learn more about NMR spectrum
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Answer:CH3COOH + NaHCO3 > H2O + CO2(g) + CH3COONa
Explanation:acid and base neutralize creating water and CO2 gas along with a salt
It will result in an increase in the rate of rxn
The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.
<h3>How do we calculate atomic mass?</h3>
Atomic mass (A) of any atom will be calculated as:
A = mass of protons + mass of neutrons
In the Thorium-234:
Number of protons = 90
Number of neutrons = 144
Mass of one proton = 1.00728 amu
Mass of one neutron = 1.00866 amu
Mass of thorium-234 = 90(1.00728) + 144(1.00866)
Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu
Given mass of thorium-234 = 234.04360 amu
Mass defect = 235.90224 - 234.04360 = 1.85864 amu
Hence required value is 1.85864 amu.
To know more about Atomic mass (A), visit the below link:
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<u>Answer:</u> The correct answer is Option b.
<u>Explanation:</u>
To calculate the amount of heat absorbed or released, we use the following equation:
.....(1)
where, q = amount of heat absorbed or released.
m = mass of the substance
c = heat capacity of water = 4.186 J/g ° C
= Change in temperature
We are given:
![m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J](https://tex.z-dn.net/?f=m%3D30g%5C%5C%5CDelta%20T%3D%5B40-0%5D%5EoC%3D40%5EoC%5C%5Cq%3D%3FJ)
Putting values in equation 1, we get:
q = 5023.2 J
We are given:
![m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J](https://tex.z-dn.net/?f=m%3D40g%5C%5C%5CDelta%20T%3D%5B40-30%5D%5EoC%3D10%5EoC%5C%5Cq%3D%3FJ)
Putting values in equation 1, we get:
q = 1674.4 J
Heat gained by Trial 1 than trial 2 = 
Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.
Thus, the correct answer is Option b.
