Answer:
-189.55°C
Explanation:
Given data:
Initial volume of gas = 3.6 L
Final volume of gas = 15.5 L
Initial temperature = ?
Final temperature = 87°C (87+273 = 360 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₁ = V₁T₂ /V₂
T₁ = 3.6 L × 360 K / 15.5 L
T₁ = 1296 L.K / 15.5 L
T₁ = 83.6 K
Kelvin to °C:
83.6 K - 273.15 = -189.55°C
Will give brainliest
Answer:
c. both gases have the same number of molecules
brainliest?
How many moles of copper are in 6,000,000 atoms of copper?
Answer: There will be 9.9632 × 10⁻¹⁸ moles of Copper in 6,000,000 atoms of Copper.
How many grams are in 2.3 x 1024 formula units of KNO3?
390 g KNO₃
General Formulas and Concepts:ChemistryAtomic Structure
Reading a Periodic TableUsing Dimensional AnalysisAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.MathPre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right Explanation:Step 1: Define
2.3 × 10²⁴ formula units KNO₃
Step 2: Identify Conversions
Avogadro's Number
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g.mol
Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
Step 3: Convert
[tex]2.3 \cdot 10^{24} \ formula \ units \ KNO_3(\frac{1 \ mol \ KNO_3}{6.022 \cdot 10^{23} \ formula \ units \ KNO_3} )(\frac{101.11 \ g \ KNO_3}{1 \ mol \ KNO_3} )[/tex] = 386.172 g KNO₃
Step 4: Check
We are given 2 sig figs. Follow sig fig rules and round.
386.172 g KNO₃ ≈ 390 g KNO₃