<h3>Answer:</h3>
64 g O₂
<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>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Divide/Multiply [Cancel Units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂
Answer:
λ = 6.5604 x 1016 nm
Explanation:
Given Data:
The energy of the red line in Hydrogen Spectra = 3.03 x 10-19
Formula to calculate Wave length
E= hv
Where E is Energy
h is Planks Constant = 6.626 x 10–34 J s
v is frequency
In turn
v= c/ λ
where c is speed of light = 3.00 x 108 m s–1
λ is wavelength = to find
Solution:
Formula to be Used:
E= hv………………………… (1)
Putting the value v in equation 1
E= h c/ λ…………………… (2)
Put the value in equation 2
3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)
By rearranging equation 3
λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J
λ = 6.5604 x 107 m
The answer is in “m”
So we have to convert it into nm
So for this to convert “m” to “nm” multiply the answer with 109
λ = 6.5604 x 107 x 109
λ = 6.5604 x 1016 nm
Answer:
- Question 19: the three are molecular compounds.
Explanation:
<em>Question 19.</em>
All of them are the combination of two kinds of different atoms in fixed proportions.
- C₂H₄: two carbon atoms per four hydrogen atoms
- HF: one hydrogen atom per one fluorine atom
- H₂O₂: two hydrogen atoms per two oxygent atoms
Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.
Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.
Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.
<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.
Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol
Water is H₂O. Its molar mass is 18.015g/mol
Calling x the number of water molecules in the hydrate, the percentage of water is:
From which we can solve for x:
Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
Bond angle of ch4 = 109.5
Answer:
Explanation:
To solve this problem, we can use the Combined Gas Laws:
Data:
p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C
p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 1 + 273.15) K = 274.15 K
T₂ = (82 + 273.15) K = 355.15 K
(b) Calculate the new pressure
