<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- Set up:
![\displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})](https://tex.z-dn.net/?f=%5Cdisplaystyle%203.60%20%5Ccdot%2010%5E%7B-5%7D%20%5C%20g%20%5C%20Au%28%5Cfrac%7B1%20%5C%20mol%20%5C%20Au%7D%7B196.97%20%5C%20g%20%5C%20Au%7D%29)
- Multiply:
![\displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au](https://tex.z-dn.net/?f=%5Cdisplaystyle%201.82769%20%5Ccdot%2010%5E%7B-7%7D%20%5C%20mol%20%5C%20Au)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
Answer:
1.
was the
value calculated by the student.
2.
was the
of ethylamine value calculated by the student.
Explanation:
1.
The
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = ![\frac{2.00 g}{180 g/mol}=0.01111 mol](https://tex.z-dn.net/?f=%5Cfrac%7B2.00%20g%7D%7B180%20g%2Fmol%7D%3D0.01111%20mol)
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = ![\frac{0.01111 mol}{0.600 L}=0.0185 M](https://tex.z-dn.net/?f=%5Cfrac%7B0.01111%20mol%7D%7B0.600%20L%7D%3D0.0185%20M)
![HAs\rightleftharpoons As^-+H^+](https://tex.z-dn.net/?f=HAs%5Crightleftharpoons%20As%5E-%2BH%5E%2B)
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:
![K_a=\frac{x\times x }{(c-x)}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7Bx%5Ctimes%20x%20%7D%7B%28c-x%29%7D)
![=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.00240%20M%5Ctimes%200.00240%20M%7D%7B%280.0185-0.00240%20%29%7D)
![K_a=3.57\times 10^{-4}](https://tex.z-dn.net/?f=K_a%3D3.57%5Ctimes%2010%5E%7B-4%7D)
was the
value calculated by the student.
2.
The
value of ethylamine = 11.87
![pH+pOH=14](https://tex.z-dn.net/?f=pH%2BpOH%3D14)
![pOH=14-11.87=2.13](https://tex.z-dn.net/?f=pOH%3D14-11.87%3D2.13)
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M
![C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-](https://tex.z-dn.net/?f=C_2H_5NH_2%2BH_2O%5Crightleftharpoons%20C_2H_5NH_3%5E%7B%2B%7D%2BOH%5E-)
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:
![K_b=\frac{x\times x}{(c-x)}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7B%28c-x%29%7D)
![=\frac{0.00741\times 0.00741}{(0.100-0.00741)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.00741%5Ctimes%200.00741%7D%7B%280.100-0.00741%29%7D)
![K_b=5.93\times 10^{-4}](https://tex.z-dn.net/?f=K_b%3D5.93%5Ctimes%2010%5E%7B-4%7D)
was the
of ethylamine value calculated by the student.
Answer:
nobmelonisegxfixcyctGkchkcigdtidtifyoc
The correct answer to this question is Water will move from left to right.
Water tends to move over to the side where there is less water.
For example,
if there's less water on the RIGHT side,
then the water will tend to move from left, to RIGHT. It <span>shows more solute molecules on the right, so water will move to this side by osmosis. I think it is to do with entropy and the tendency for systems to move to equilibrium if there is an increase in entropy</span>
The answer is D, because gems are usually not mettallic