Answer:
The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.
Explanation:
Given conjugate base CN⁻ => weak acid => HCN => Ka =4.9 x 10⁻¹⁰
Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴
Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻
Answer:
0.0187 M
Explanation:
Step 1: Write the balanced neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
18.7 mL of 0.01500 M HCl react.
0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol
Step 3: Calculate the reacting moles of NaOH
The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.
Step 4: Calculate the molarity of NaOH
2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.
[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M
Answer:
P₅O₁₂
<em>Explanation: </em>
Assume that you have 100 g of the compound.
Then you have 44.7 g P and 55.3 g O.
1. Calculate the <em>moles</em> of each atom
Moles of P = 44.7 × 1/30.97 = 1.443 mol Al
Moles of O = 55.3 × 1/16.00 = 3.456 mol O
2. Calculate the <em>molar ratios</em>.
P: 1.443/1.443 = 1
O: 3.456/1.443 = 2.395
3. Multiply by a number to make the ratio close to an integer
P: 5 × 1 = 5
O: 5 × 2.395 = 11.97
3. Determine the <em>empirical formula
</em>
Round off all numbers to the closest integer.
P: 5
O: 12
The empirical formula is <em>P₅O₁₂</em>.
