<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer: The empirical formula for the given compound is 
Explanation : Given,
Mass of O = 0.370 g
Mass of N = 0.130 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Oxygen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.
For Oxygen = 
For Nitrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of O : N = 2 : 1
Hence, the empirical formula for the given compound is 
The correct answer is 0.011330374846536 Hope this helps ;D
Answer:
Helium, this noble gas has the least density at STP
Explanation:
Density = mass / volume
= 69g / 23 ml
= 3 g / ml.
Thus, the density of the sample is 3 grams per ml or 3g/ ml