<h3>
Answer:</h3>
3.33 mol 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>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
60.0 g H₂O (Water)
<u>Step 2: Identify Conversions</u>
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: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Convert</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
3.32963 mol H₂O ≈ 3.33 mol H₂O
<span>The carbon sp2 is hybridized in any cabozxylic acid oninon and it forms one simga bond with each oxygine and one sigma bond with carbon backbone.</span>
Rydberg Eqn is given as:
1/λ = R [1/n1^2 - 1/n2^2]
<span>Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m </span>
<span>R is the Rydberg constant: R = 1.09737×10^7 m-1 </span>
<span>From Brackett series n1 = 4 </span>
<span>Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] </span>
<span>Some rearranging and collecting up terms: </span>
<span>1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] </span>
<span>1= 28.82[1/16 – 1/n2^2] </span>
<span>28.82/n^2 = 1.8011 – 1 = 0.8011 </span>
<span>n^2 = 28.82/0.8011 = 35.98 </span>
<span>n = √(35.98) = 6</span>
Answer:Specific heat capacity at Constant volume of Nitrogen =742.32JKg.K
Explanation:
The Specific heat capacity at Constant volume of an ideal gas is given as
c =Cv / M
Cv= Constant volue of gas
M= Molar mass
But First, we determine the Constant volume, Cv which is given as
Cv = 5/2R since Nitrogen is a diatomic gas, N2 where R= 8.314Jmol.k
= 5/2 x 8.314Jmol.K
=20.785Jmol.K
Specific heat capacity at Constant volume, c
c= Cv/M
M=molar mass = 28.0 g/mol.
changing to kg/mol =28/1000= 0.028kg/mol
Therefore c =20.785Jmol.K/0.028kg/mol
=742.32JKg.K
