Answers
By decomposing 157g of NaCl, approximately 1.35 moles of chlorine can be produced.
To determine the number of moles of chlorine that could be produced by decomposing 157g of NaCl, we need to use the molar mass of NaCl and apply stoichiometry.
The molar mass of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl), which is approximately 22.99 g/mol + 35.45 g/mol = 58.44 g/mol.
Now, we can calculate the number of moles of NaCl:
Moles of NaCl = Mass of NaCl / Molar mass of NaCl
Moles of NaCl = 157 g / 58.44 g/mol
Moles of NaCl ≈ 2.69 mol
According to the balanced equation 2NaCl → 2Na + Cl₂, we can see that for every 2 moles of NaCl, we produce 1 mole of Cl₂.
Therefore, using the stoichiometric ratio, we can calculate the number of moles of chlorine produced:
Moles of Cl₂ = (Moles of NaCl / 2) × 1
Moles of Cl₂ = 2.69 mol / 2
Moles of Cl₂ ≈ 1.35 mol
Thus, by decomposing 157g of NaCl, approximately 1.35 moles of chlorine can be produced.
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b
Carbon is the cation and oxygen is the anion.
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Answer:
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Answer: 1 more of hydrogen =22.4dm3
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Answer: B.
Explanation:
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Explanation:
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Answer:
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The N2O decomposition, 2 N2O (g) → 2 N2 (g) + O2 (g), is a zero order reaction that has a reaction rate constant of 2.00 × 10−3 M/s. Initially, 2.00 moles of N2O gas are infused into a 1.00 L vessel at 300. K.Calculate the pressure inside the vessel at 100. seconds after the start of the reaction. Temperature is held constant.
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Answer:
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Answer:
1) during a phase change: particles overcome forces of attraction and temperature stays the same not during a phase change: temperature rises 2)Particle motion decreases, and electrostatic forces pull particles closer together.
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Answer:
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Answer:
because those are the only thermometers that are truly accurate for kelvins
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Answer:
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Answer:
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The atomic mass of boron is 10.81 u.
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To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
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11u−ux=10.81u
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Answers
Answer:
sensory experience of a chemical reaction called combustion. In a way, fire is like the leaves changing color in fall, the smell of fruit as it ripens, or a firefly's blinking light. All of these are sensory clues that a chemical reaction is taking place
Explanation:
Sensory encounter with the combustion chemical process. Fire can be compared to the fall foliage's shifting hues, the aroma of ripe fruit, or the blinking light of fireflies. These are all sensory indicators that a chemical reaction is occurring.
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At a certain temperature, the equilibrium pressures of NO2 and N2O4 are 1.4 bar and 0.46 bar, respectively. If the volume of the container is doubled at constant temperature, what would be the partial pressures of the gases when equilibrium is re-established?
Answers
The partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.
Let x be the mole fraction of NO₂ and x' be the mole fraction of N₂O₄.
The total pressure according to Dalton's law of partial pressure is the sum of the partial pressures of each gas.
Let P be the total pressure of the gases, P' be the partial pressure of NO₂ = 1.4 bar and P" be the partial pressure of N₂O₄ = 0.46 bar.
So, P = P' + P"
= 1.4 bar + 0.46 bar
= 1.86 bar
Since P' = xP
x = P'/P
= 1.4 bar/1.86 bar
= 0.753
Also, P" = xP
x' = P"/P
= 0.46 bar/1.86 bar
= 0.247
Now, since the volume of the container is doubled at constant temperature, we use Boyle's law to determine the new pressure. P₁.
Boyle's law states that the pressure of a given mass of gas is inversely proportional to its volume provided the temperature remains constant. It is written mathematicaly as PV = constant
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P₁ = (V/V₁)P
Since V/V₁ = 1/2
P₁ = (V/V₁)P
P₁ = P/2
P₁ = 1.86/2 bar
P₁ = 0.93 bar
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= 0.753 × 0.93 bar
= 0.70 bar
The new partial pressure of N₂O₄, P₃ = x'P₁
= 0.247 × 0.93 bar
= 0.23 bar
So, the partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.
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