Answer:
What I came up with was D equaling out to be 42
Explanation:
Im assuming you just do the math
A. 3.5 x 104 = 364
B. 2.4 x 103 = 247.2
C. 1.4 x 10^2_2 = 280
D. 4.9 x 10-7 = 42
I hope I helped.
Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:

Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:



<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
<h3>
Answer:</h3>
1.43 × 10⁻²⁰ mol Li
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
8.63 × 10³ atoms Li
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li
Mr: 207.2
m=n×Mr= 6.53×207.2= 1353.02g
Three classes: 28×3=84 students
8 test tubes per student: 84× 8= 672
hope this helps!