Chemical and physical properties of the seaborgium element
1 answer:
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Answer:
7.32g of HNO3 are required.
Explanation:
1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.
From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:
• starting with the 4.30 grams of Ca(OH)2.
,• using the molar mass of Ca(OH)2 (74g/mol).
,• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .
,• using the molar mass of HNO3 (63.02g/mol).
So, 7.32g of HNO3 are required.
<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
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<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