Answer:
We have to add 9.82 grams of calcium acetate
Explanation:
Step 1: Data given
Molarity of the calcium acetate solution = 0.207 M
Volume = 300 mL = 0.300 L
Molar mass calcium acetate = 158.17 g/mol
Step 2: Calculate moles calcium acetate
Moles calcium acetate = molarity * volume
Moles calcium acetate = 0.207 M * 0.300 L
Moles calcium acetate = 0.0621 moles
Step 3: Calculate mass calcium acetate
Mass calcium acetate = moles * molar mass
Mass calcium acetate = 0.0621 moles * 158.17 g/mol
Mass calcium acetate = 9.82 grams
We have to add 9.82 grams of calcium acetate
Step 1: Write the unbalanced equation,
C₂H₆ + O₂ → CO₂ + H₂<span>O
There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,
C₂H₆ + O₂ → 2 CO₂ + H₂O
Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.
C₂H₆ + O₂ → 2 CO₂ + 3 H₂O
At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.
C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O
Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.
( C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O ) × 2
2 C₂H₆ + 3 O₂ → 4 CO₂ + 6 H₂O
I’m pretty sure it’s true
In the question, the number of atoms per unit cell is required for:
A) Polonium (Po)
In polonium, the structure is simple cubic, meaning there are 8 corner atoms, which add up to one atom per unit cell.
B) Manganese (Mn)
The structure of the Mn can be considered to be a body centered cubic (BCC) and the number of atoms for this is 8 corner atoms and 1 central atoms, making a total of 2 atoms per unit cell.
C) Silver (Ag)
Silver has a face centered cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.
