Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2
Spent fuel that can no longer be used to create energy is waste.
Answer:
C3 H6 O2
Explanation:
first divide their mass by their respective molar mass, we get:
30.4 moles of C
61.2 moles of H
20.25 moles of O
now divide everyone by the smallest one of them then we get
C= 1.5
H= 3
O= 1
since our answer of C is not near to any whole number so we will multiply all of them by 2
so,
C3 H6 O2 is our answer
The average mass of an atom is calculated with the formula:
average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on
For the boron we have two isotopes, so the formula will become:
average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)
We plug in the values:
10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)
10.81 = 1.98 + 0.8020 × mass of isotope (2)
10.81 - 1.98 = 0.8020 × mass of isotope (2)
8.83 = 0.8020 × mass of isotope (2)
mass of isotope (2) = 8.83 / 0.8020
mass of isotope (2) = 11.009975
mass of isotope (1) = 10.012938 (given by the question)
Answer:
Density of the object is 8759.494 grams/L
Explanation:
As we know density of an object is mass of the object divided by its volume
Given
Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.
Hence, volume of object is equal to
mL
Mass of the object is
grams
Density
grams/L
Density of the object is 8759.494 grams/L