Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
===============
Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
==============
Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
Answer:
Cabon-12 has same average atomic weight and mass number.
Explanation:
carbon-12 has average atomic weight 12 amu and mass number of 12.
amu represents average of mass of a nucleon.
As carbon-12 has same average atomic mass and mass number therefore carbon-12 is a good standard to determine average mass of a nucleon.
Again, abundance of carbon-12 isotope is almost equal to 99%. Therefore fluctuation of average atomic weight from 12 amu is very very low.
So, carbon-12 is taken as a standard to determine mass of a nucleon.
Hence atomic mass of carbon-12 is 12 amu.
Answer: An atom can be considered unstable in one of two ways. If it picks up or loses an electron, it becomes electrically charged and highly reactive. Such electrically charged atoms are known as ions. Instability can also occur in the nucleus when the number of protons and neutrons is unbalanced.
Explanation:
Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K
The correct answer is 13 I just did it bro
