<span>1. What is the molar mass of gold?
Molar mass is a unit that expresses the mass of a molecule per one mol. The molar mass can be obtained by adding the neutron with the proton of the atoms. Gold has atomic number 79 so the proton is 79. The number of the neutron is 118. Then the molar mass would be: 79 + 118 = </span>197 g/mol<span>
</span><span>2. Calculate the number of moles of gold (Au) in the sample. Show your work.
</span>In this question, you are given the mass of the gold and asked for how many moles the sample has. To find the number of moles you just need to divide the weight by the molar mass.
For 45.39 grams of gold, the number of moles would be:
45.39 / (197g/mol)= 0.23 moles
3. Calculate the number of atoms of gold (Au) in the sample. Show your work.Moles is unit of a number of molecules but 1 mol doesn't represent 1 molecule. The number of atoms can be obtained by multiplying the number of moles with Avogadro number. The calculation would be:
0.23 moles * (6.023 * 10^23 molecules/mol)= 1.387 * 10^23 molecules
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:
kJ/mol
Explanation:
Given and known facts
Mass of Benzene
grams
Mass of water
grams
Standard heat capacity of water
J/g∙°C
Change in temperature ΔT
°C
Heat

Heat released by benzine is - 7.82 kJ
Now, we know that
grams of benzene release
kJ heat
So,
g benzine releases

kJ/g
mol C6H6
Heat released

kJ/mol
Answer:
Kp = 0.022
Explanation:
<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>
<em />
The equilibrium of ammonia occurs as follows:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where Kp is defined as:

<em>Where P represents partial pressure of each gas.</em>
<em />
As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:
P(NH₃) = 2.3atm - 2X
<em>Where X represents reaction coordinate</em>
<em />
Thus, pressure of hydrogen and nitrogen is:
P(N₂) = X
P(H₂) = 3X.
As partial pressure of hydrogen is 0.69atm:
3X = 0.69
X = 0.23atm:
P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm
P(N₂) = 0.23atm
P(H₂) = 0.69atm

<h3>Kp = 0.022</h3>