Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is 
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- 
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
Answer: A
Explanation: Beta particles have a charge of -1
Answer:
The answer to your question is 0.4 moles of Oxygen
Explanation:
Data
Octane (C₈H₈)
Oxygen (O₂)
Carbon dioxide (CO₂)
Water (H₂O)
moles of water = ?
moles of Oxygen = 1
Balanced chemical reaction
C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O
Reactant Element Products
8 C 8
8 H 8
20 O 20
Use proportions to solve this problem
10 moles of Oxygen ----------------- 4 moles of water
1 mol of Oxygen ------------------ x
x = (4 x 1) / 10
x = 4 / 10
x = 0.4 moles of water
Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
