5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.
Answer:
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Answer:
The atom economy of ethane in this process is 19.72 %.
What is atom economy?
The conversion efficiency of a chemical reaction in terms of all the atoms involved and the desired products produced is known as atom economy (atom efficiency/percentage).
Explanation:
C₁₀H₂₂ → C₈H₁₈ + C₂H₄
Molecular weight of C₁₀H₂₂ = 142.28
Molecular weight of C₈H₁₈ = 114.228
Molecular weight of C₂H₄ = 28.053
% Atom economy = 
= 
= 19.716 %
≈ 19.72 %
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Answer:
The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl
Explanation:
Step 1: Data given
pH of a buffer = pKa + log ([A-]/[Ha])
a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)
pH = -log( 1.8 * 10^-5) + log (1/1)
pH = -log( 1.8 * 10^-5)
pH = 4.74
a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)
pH = -log( 4.9 * 10^-10) + log (1/1)
pH = -log( 1.8 * 10^-5)
pH = 9.30
a mixture of 1.0 M HCl and 1.0 M NaCl
The solution made from NaCl and HCl will NOT act as a buffer.
HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.
a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)
Ka * Kb = 1*10^-14
Ka = 10^-14 / 1.76*10^-5
Ka = 5.68*10^-10
pH = -log( 5.68*10^-10) + log (1/1)
pH = -log( 5.68*10^-10)
pH = 9.25
The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl
