Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
Bases have a Ph do gretater than 7
They are bitter in taste
They are soapy texture
Sulfur trioxide (SO3) is a chemical compound that is a significant pollutant in gaseous form as it is involved in the production of acid rain.
Industrially, sulfur trioxide is an important precursor to sulfuric acid and is formed from the reaction between sulfur dioxide (SO2) and oxygen gas (O2) as shown in the chemical equation below.
1 is D - double-replacements do not make solid metals
2 is A - to have complete combustion the original compound must ONLY have C, H and O
3 is B - the elemental Mg replaces the H in the HCl
The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
