Objects and substances in motion is I

A sub-shell with n = 6, l = 2 can accommodate a maximum of:

Mrrafil [7]

Answer:

72

Explanation:

2n^2

n=6

2(6)^2=2×36

=72

8 0

3 years ago

Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen

Oksanka [162]

<h3>Answer:</h3>

10.6 mol NO

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O

[Given] 13.2 mol O₂

Step 2: Identify Conversions

[RxN] 5 mol O₂ → 4 mol NO

Step 3: Stoich

[DA] Set up: $\displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 10.56 \ mol \ NO$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

10.56 mol NO ≈ 10.6 mol NO

5 0

3 years ago

A 2,000g of C-14 is left to decay radioactively. The half-life of Carbon-14 is approximately 5,700 years. What percentage of the

dsp73

Answer is: 12,6% (1/8) percentage of the sample will remain.
c₀ - initial amount of C-14.
c - amount of C-14 remaining at time.
t = 5700 y.
First calculate the radioactive decay rate constant λ:
λ = 0,693 ÷ t = 0,693 ÷ 5700 y = 0,000121 1/y = 1,21·10⁻⁴ y.
c = c₀·e∧-λ·t.
c = 2000 · e∧-(0,000121 1/y · 17100 y).
c = 252 g.
ω = 252 g ÷ 2000 g = 0,126 = 12,6%.

4 0

3 years ago

Read 2 more answers

Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains

erastova [34]

Given that:

The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
volume = 250 mL = 0.250 L
Mass of butane = 0.8 g
the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

$\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}$