Which of the following measurements has three significant figures?
2 answers:
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<h3>Answer:</h3>
1.2 × 10⁻⁸ mol Pb
<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> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[Given] 7.2 × 10¹⁵ atoms Pb
<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>
1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb
19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.
<h3>What is vapour pressure?</h3>Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.
Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2
Volume of (V) = 40 liter
Temperature (T) = 22°C = 22 + 273 = 295 K
Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm
Moles of (n) can be calculated by ideal gas equation.
PV = nRT
n = 1.007 40 ÷ 0.0821 295 = 1.663
Balanced chemical reaction;
2 + 13
---> 8
+ 10
From reaction;
13 moles require 2 moles
So, 1.663 moles will require = 2 x 1.663 ÷13 = 0.256 moles of
Thus is a limiting reagent. So it will drive the yield of
.
Moles of produced = (8/2) 0.2 = 0.8 moles
Pressure of (P) = 102 - 2.24 = 99.76 kPa = 99.76 ÷ 101.325 = 0.985 atm
Applying the ideal gas equation for ,
PV = nRT
0.985 V = 0.8 0.0821 x 295
V = 19.7 liter
The volume of produced = 19.7 liter.
Learn more about the vapour pressure here:
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