B A C ( Blood Alcohol Content ) of 0.10 means that there are 0.10 g of alcohol for every dl of blood.
5 L = 50 dl
50 * 0.10 g = 5 g
In his blood is circulating 5 grams of alcohol.
1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3
2. Na₂CO₃ as a limiting reactant
<h3>Further explanation</h3>
Given
Reaction
2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃
Required
mol ratio
Limiting reactant
Solution
The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)
1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3
2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67
Na₂CO₃ as a limiting reactant (smaller)
Answer:
combination of a very light gas (hydrogen) and a substance that is found in all living things (carbon). Hydrocarbons are found in petrol, coal and natural gas
<h3>
Answer:</h3>
3.3 × 10²³ molecules Cu(NO₃)₂
<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>
0.55 mol Cu(NO₃)₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂
Answer is: [COCl₂] > [CO][Cl₂]
Chemical reaction: COCl₂(g) ⇄ CO(g) + Cl₂(g); Keq = 8.1 x 10⁻⁴.
Keq = [CO] · [Cl₂] / [COCl₂]; equilibrrium constant of chemical reaction.
[CO] · [Cl₂] / [COCl₂] = 0,00081.
Equilibrium product concentration is much more less than equilibrium concentration of reactant.