Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
4 carbon electrons. Hope this helps have a nice day
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = <em>0.0385 atm</em> → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = <em>0.7515atm</em>
O₂ = 0.210atm - X = <em>0.1715atm</em>
NO = 2X = <em>0.0770atm</em>
Answer: Uranium-235.
Radioactive isotopes are used to determine the age of antique objects, including fossils.
The half-life time of the radioactive elements is what permits the process of dating.
The half-life of C-14 is too short to be useful to date too old objects.
Precambrian time is the most antique era. C-14 hal-life is about 5730 years and Precambrian time is millions or billions of years ago. Given that the hal-life of U-235 is 704 million years it is appropiate to date the fossils from the Precambrian era.
<h2>
Answer: 131.9 g</h2>
<h3>
Explanation:</h3>
<u>Write a Balanced Equation for the decomposition</u>
CaCO₃ → CaO + CO₂
<u></u>
<u>Find Moles of CO₂ Produced</u>
Since the mole ratio of CaCO₃ to CO₂ is 1 to 1,
the moles of CaCO₃ = moles of CO₂
moles of CaCO₃ = mass ÷ molar mass
= 300 g ÷ 100.087 g/mol
= 2.997 moles
∴ moles of CO₂ = 2.997 moles
<u>Determine Mass of CO₂</u>
Mass = moles × molar mass
= 2.997 mol × 44.01 g/mol
= 131.9 g
<u></u>
<h3>∴ when 300 g of calcium carbonate is decomposed, it produces 131.9 g of carbon dioxide.</h3>
