1) Convert 12.9 liters of Oxygen to mol at the given conditions:
PV = nRT ⇒ n = PV/RT
n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]
n = 0.636 mol of O2
2) use the stoichiometry derived from the balanced chemical equation
1mol C2H4 / 3 mol O2 = x mol C2H4 / 0.636 mol O2
x = 0.636 / 3 mol O2 = 0.212 mol O2.
Answer: 0.212 mol O2
Answer:
Approximately 2000 J.
General Formulas and Concepts:
<u>Thermodynamics</u>
Specific Heat Formula: q = mcΔT
- <em>q</em> is heat (in J)
- <em>m</em> is mass (in g)
- <em>c</em> is specific heat (in J/g °C)
- ΔT is change in temperature (in °C or K)
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>c</em> = 0.897 J/g °C
[Given] <em>m</em> = 79 g
[Given] ΔT = 28°C
[Solve] <em>q</em>
<em />
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (79 g)(0.897 J/g °C)(28 °C)
- Multiply [Cancel out units]: q = (70.863 J/°C)(28 °C)
- Multiply [Cancel out units]: q = 1984.16 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
1984.16 J ≈ 2000 J
Answer:
Box is made up of <em>copper</em>, because density is <em>8.96 g/cm³.</em>
Explanation:
Given data:
Volume of box = 17.63 cm³
Mass of box = 158 g
Which metal box is this = ?
Solution:
First we will calculate the density of box then we will compare it with the density value of given metals.
d = m/v
d = 158 g/ 17.63 cm³
d = 8.96 g/cm³
The calculated density is similar to the given density value of copper thus box is made up of copper.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M