<h3>
Answer:</h3>
7.3 × 10⁻⁷ g Ni
<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
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
7.5 × 10¹⁵ atoms Ni
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Ni - 58.69 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
7.30945 × 10⁻⁷ g Ni ≈ 7.3 × 10⁻⁷ g Ni
The question is incomplete, here is the complete question:
Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴
a) 2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
<u>Answer:</u> The pH of the solution is 3.546
<u>Explanation:</u>
We are given:
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L
To calculate the molarity of solution, we use the equation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5BHCOONa%5D%7D%7B%5BHCOOH%5D%7D%29)
= negative logarithm of acid dissociation constant of formic acid = 3.75
![[HCOONa]=\frac{0.230}{1}](https://tex.z-dn.net/?f=%5BHCOONa%5D%3D%5Cfrac%7B0.230%7D%7B1%7D)
![[HCOOH]=\frac{0.370}{1}](https://tex.z-dn.net/?f=%5BHCOOH%5D%3D%5Cfrac%7B0.370%7D%7B1%7D)
pH = ?
Putting values in above equation, we get:
Hence, the pH of the solution is 3.546
Im not completely sure but It might be Cycling of Matter
The answer cl atoms are in zn(cio3)2 is A:2 because I looked it up and it gave me 2.
Answer:
1.0 *10^(-4) mol
Explanation:
For gases:
n1/n2 = V1/V2
n1/3.8*10^(-4) mol = 230 mL/ 860 mL
n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol
