<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
Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
Answer:
Exam 3 Material
Homework Page Without Visible Answers
This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
Explanation:
Molarity = mol/liter
0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L
For every liter there is 1000 mL:
.138L • 1000mL =138mL KOH
Answer:
Mass = 0.37 g
Explanation:
Given data:
Number of moles of sulfur = 11.9 mol
Mass of sulfur in 11.9 mol = ?
Molar mass of sulfur = 32.06 g
Solution:
Number of moles = mass/molar mass
by putting values,
11.9 mol = mass/ 32.06 g/mol
Mass = 11.9 mol × 32.06 g/mol
Mass = 0.37 g
