The limiting reactant can be determined by calculating the
moles supplied / moles stoich ratio and the lowest is the limiting reactant.
Fe(CO)5 ratio = [6 g / 195.9 g/mol] / 1
Fe(CO)5 ratio = 0.0306
PF3 ratio = [4 g / 87.97 g/mol] / 2
PF3 ratio = 0.0227
H2 ratio = [4 g / 2 g/mol] / 1
H2 ratio = 2
<span>We can see that PF3 has the lowest ratio, so it is the
limiting reactant.</span>
Answer:
1) 0.009 61 g C; 2) 0.008 00 mol C
Step-by-step explanation:
You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.
M_r: 12.01 44.01
C + ½O₂ ⟶ CO₂
m/g: 0.352
1) <em>Mass of C
</em>
Convert grams of CO₂ to grams of C
44.01 g CO₂ = 12.01 g C
Mass of C = 0.352 g CO₂ × 12.01 g C/44.01 g CO₂
Mass of C = 0.009 61 g C
2) <em>Moles of C
</em>
Convert mass of C to moles of C.
1 mol C = 12.01 g C
Moles of C = 0.00961 g C × (1 mol C/12.01 g C)
Moles of C = 0.008 00 mol C
All the carbon comes from Compound A, so there are 0.008 00 mol C in Compound A.
<span>The fog in the mirror is the condensation of water vapor as it touches a colder surface. When you are running cold water you just cool down everything around it. Now the vapor coming from the hot shower will mostly condense right there and will not reach the mirror.</span>
Answer: A) Bent or angular, polar
Explanation:
The central atom oxygen has two lone pairs and two bond pairs in 
. The number of electron pairs are 4, that means the hybridization will be 
and the electronic geometry of the molecule will be tetrahedral. But as there are two lone pair of atoms around the central oxygen atom, repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent shape.
The compound is polar as the net dipole moment of oxygen - fluoride bonds do not cancel each other out.
<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
We are given that;
1 stadium holds = 1 × 10^5 people
Number of iron atoms is 1 × 10^18 atoms
Assuming the stadium would carry an equivalent number of atoms as people.
Then, 1 stadium will carry 1 × 10^5 atoms
Therefore,
To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.
Number of stadiums = Total number of atoms ÷ Number of atoms per stadium
= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium
= 1 × 10^13 Stadiums
Thus, 1 × 10^18 atoms would occupy 1 × 10^13 stadiums
