1 mole -------- 22,4 L ( at STP )
0.030 moles --- ?
V = 0.030 * 22,4 / 1
V = 0.672 L
hope this helps!
Answer:
7.640 kg
Explanation:
Step 1: Write the balanced complete combustion equation for ethanol
C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O
Step 2: Calculate the moles corresponding to 4 kg (4000 g) of C₂H₆O
The molar mass of C₂H₆O is 46.07 g/mol.
4000 g × 1 mol/46.07 g = 86.82 mol
Step 3: Calculate the moles of CO₂ released
86.82 mol C₂H₆O × 2 mol CO₂/1 mol C₂H₆O = 173.6 mol CO₂
Step 4: Calculate the mass corresponding to 173.6 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
173.6 mol × 44.01 g/mol = 7640 g = 7.640 kg
4.2 × 10²² atoms Al × (1 mol Al / 6.022 × 10²³) = moles Al
Last one: fraction 1 mole Al over 6.022 × 10²³ atoms Al
Answer:
35
Explanation:
The number of protons in bromine-80 is 35, it is the same with the atomic number.
1. Pipette, measuring cylinder
