<h3>
Answer:</h3>
2.0 mol C₆H₁₂O₆
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:

- Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆
Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka =
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.
Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.
Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.
Using the formula;
k = 0.693/t1/2
k = 0.693/21 hours = 0.033 hr-1
Using;
N=Noe^-kt
N = mass of radioactive sample at time t
No = mass of radioactive sample initially present
k = decay constant
t = time taken
Substituting values;
2.5/320= e^- 0.033 t
0.0078 = e^- 0.033 t
ln (0.0078) = 0.033 t
t = ln (0.0078)/-0.033
t = 147 hours
Learn more: brainly.com/question/6111443
Answer:
Hydrogen and Cobalt
Explanation:
Break up each individual element
- Hope that helps! Please let me know if you need further explanation.