Answer:
A titration
Explanation:
A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.
By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.
From the chemical formula of sulfuric acid, we can see the molar ratio:
H : S : O
2 : 1 : 4
Now, we convert the mass of hydrogen given into the moles of hydrogen. This is done using
Moles = mass / Mr
Moles = 7.27 / 1
Moles = 7.27
Therefore, the moles will be:
S = 7.27 / 2 = 3.64 moles
O = 7.27 * 2 = 14.54 moles
Now, the respective masses are:
S = 32 * 3.64 = 116.48 grams
O = 16 * 14.54 = 232.64 grams
Arkeisha is correct because the fluid in an alkaline battery has a ph between 7.1 and 14.0
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
