Frequency= 1/period
hence ans= 1/0.25
= 4s
Explanation:
The answer is 0.5 moles of gold
Answer:
V O2 = 1.623 L
Explanation:
- 1 mol ≡ 6.022 E23 molecules
∴ molecules O2 = 4.00 E22 molecules
⇒ moles O2 = (4.00 E22 molecules O2)×(mol O2/6.022 E23 molecules)
⇒ moles O2 = 0.0664 moles
at STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
assuming ideal gas:
∴ V = RTn/P
⇒ V O2 = ((0.082 atm.L/K.mol)(298 K)(0.0664 mol))/( 1 atm)
⇒ V O2 = 1.623 L
Answer:7,070.74
Explanation:because frequency has a lot of energy
Answer: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Explanation:
