Thus problem is providing us with the mass of iron (III) oxide as 12.4 g so the moles are required and found to be 0.0776 mol after the calculations:
<h3>Mole-mass relationships:</h3>
In chemistry, we use mole-mass relationships in order to calculate grams from moles and vice versa. In this case, since we are given the mass of iron (III) oxide as 12.4 g one can calculate the moles by firstly quantifying its molar mass:
Then, we prepare a conversion factor in order to cancel out the grams and thus, get moles:
Learn more about mole-mass relationships: brainly.com/question/18311376
Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
Answer:- B: 
is the right answer.
Solution:- The balanced equation is:
We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.
From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.
So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.
The complete set up looks as:
=
Hence, the right option is B: .
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
One chemical change would be baking the muffin.
One physical change would be chopping the fruits to add to the mixture
