I honestly don’t know I just need points sorry
1 mole of any gas occupies a volume of 22.4L, therefore there are 3.022×10^24 molecules of oxygen gas in 113.97 L
PH is the test of acidity or basicity of a solution. it follows the formula:
pH = pKa + log [salt] / [acid] where NaF is the salt and HF is the acid in this case.
By literature, Ka of HF is 3.5*10^-4
<span>pKa= -log(Ka)=</span><span> 3.46 </span>
<span>pH = pKa + log [NaF / [HF] </span>
4.05 = 3.46 + log [NaF / [HF]
log [NaF / [HF]<span> = 0.59
</span>
[NaF / [HF] = 3.89
Answer:
m AgCl = 28.395 g
Explanation:
- AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
∴ [AgNO3] = 0.156 mol/L
∴ V = 1.27L
⇒ mol AgNO3 = 0.156 mol/L * 1.27 L = 0.19812 mol AgNO3
mass AgCL:
⇒ m AgCl = 0.19812 mol AgNO3 * mol AgCl/molAgNO3 * 143.32gAgCl/molAgCl
⇒ m AgCl = 28.395 g
Answer:
0.19 mol H₂O
General Formulas and Concepts:
<u>Chem</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
Given: 3.5 g H₂O from RxN
<u>Step 2: Define conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
<u /> = 0.194229 mol H₂O
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
0.194229 mol H₂O ≈ 0.19 mol H₂O