Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptur
1 answer:
Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.
Hank's Garage has an air compressor with a holding tank that contains a volume of 200L (V₁) of compressed air at a pressure of 5200 torr (P₁).
One day a hose ruptured and all the compressed air was released. The final pressure was the atmospheric pressure (1 atm = 760 torr) (P₂).
We can calculate the new volume (V₂) in these conditions using Boyle's law, which states there is an inverse relationship between the volume and the pressure of an ideal gas.
Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.
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1.5 % is the percent yield in the reaction.
Explanation:
Given that:
original mass of the sample used in reaction = 1.897 grams
product formed after decomposition = 1.071 grams
The reaction for the decomposition:
Mg(HCO3)2 (s) ⇒ CO2 (g) + H2O (g) + MgCO2 (s)
It says that 1 mole of Mg(HCO3)2 yielded 1 mole of MgCO2 on decomposition
68.31 grams/mole or 68.31 grams of MgCO2 is formed
percent yield = x 100
putting the values in the equation:
percent yield =
= 0.015 x100
PERCENT YIELD = 1.5 %
Answer:
4.18 g
Explanation:
The formula for the calculation of moles is shown below:
Given: For Li
Given mass = 2.50 g
Molar mass of Li = 6.94 g/mol
<u>Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles</u>
Given: For
Given mass = 2.50 g
Molar mass of = 28.02 g/mol
<u>Moles of = 2.50 g / 28.02 g/mol = 0.08924 moles</u>
According to the given reaction:
6 moles of Li react with 1 mole of
1 mole of Li react with 1/6 mole of
0.3602 mole of Li react with mole of
Moles of that will react = 0.06 moles
Available moles of = 0.08924 moles
is in large excess. (0.08924 > 0.06)
Limiting reagent is the one which is present in small amount. Thus,
Li is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
6 moles of Li gives 2 mole of
1 mole of Li gives 2/6 mole of
0.3602 mole of Li react with mole of
Moles of = 0.12
Molar mass of = 34.83 g/mol
Mass of = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g
<u>Theoretical yield = 4.18 g</u>
<h3>Answer:</h3>
2.04 mol CBr₄
<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>Organic</u>
- Writing Organic Compounds
- Writing Covalent Compounds
- Organic Prefixes
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
675 g CBr₄
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of Br - 79.90 g/mol
Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol
<u>Step 3: Convert</u>
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<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄