Answer:
2812.6 g of H₂SO₄
Explanation:
From the question given above, the following data were obtained:
Mole of H₂SO₄ = 28.7 moles
Mass of H₂SO₄ =?
Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:
Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)
= 2 + 32 + 64
= 98 g/mol
Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:
Mole of H₂SO₄ = 28.7 moles
Molar mass of H₂SO₄ =
Mass of H₂SO₄ =?
Mole = mass / Molar mass
28.7 = Mass of H₂SO₄ / 98
Cross multiply
Mass of H₂SO₄ = 28.7 × 98
Mass of H₂SO₄ = 2812.6 g
Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄
Answer:
Rate will double
Explanation:
Since the concentration of A is doubling the rate of reaction will also double, since in this scenario concentration is proportional to rate of reaction. So if the concentration of A were to triple the rate of reaction would also triple.
Molarity is moles divided by liters so do .732 divided by .975 liters.
<h3>
Answer:</h3>
16.7 g H₂O
<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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
To find them you would have numbers of the elements in percentage or grams then you divide them by their molar mass to get their moles. From there you divide by the smallest number. Round it to two or one sig fig. If you have a number that is for ex. 2.5 you multiply it by 2 to make it whole as well the other whole numbers. Then to find the molecular formula the problem must give you another molar mass and using your empirical formula convert it to its molar mass then you divide them, larger number over smaller number. You should get a number round it to 1 sig fig. Now you use that number and multiply the subscripts on the empirical formula to get the molecular formula.
