The cure for polio is an applied science. The vaccination as the cure of the disease uses applied science<span> that applies existing knowledge in science to develop practical technology or inventions like the vaccine. Answer to this problem is applied/</span>
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Answer:
Blank #1: Polyatomic
Blank #2: 2 (see explanation)
Blank #3: 1 (see explanation)
Explanation
Both the ammonium ion and the sulfate ion contain more than one atom in each ion. The two species are thus polyatomic. The chloride ion , for example, is monoatomic.
Superscripts above formulae of the ions indicate their charge. Each ammonium ion carries a positive one (+1) charge. Each sulfate ion carries a charge of negative two (-2).
Ammonium sulfate is an ionic compound. A sample of this compound contain myriads of ammonium ions and sulfate ions. The ions are packed in three-dimensional lattices. Thus unlike water, ammonium sulfate does not exist as molecules in nature.
Assuming that the second and third blanks refers to a formula unit, rather than a molecule, of ammonium sulfate. The empirical formula of ammonium sulfate gives the minimum whole-number ratio between the two ions in a sample.
Charges shall balance between the two ions. Ammonium ions are of charge +1. Sulfate ions are of charge -2. The sample shall thus contain two ammonium ions for every one sulfate ion.
The empirical formula of ammonium sulfate is therefore .
There are thus two ammonium ions and one sulfate ion in each formula unit of ammonium sulfate.
Answer:
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Explanation:
Answer:
(a) m = 50.916 g
(b) CO2 is the limiting reagent and C is the reagent in excess.
(c) the mass excess left = 4.084 g
Explanation:
Balance the given equation first:
(a) Given:
mass of CO2 = 40.0 g
mass of C = 15.0 g
mass of CO = ?
To find the mass of CO that will be produced, we need to find the limiting reactant first. To find the limiting reactant we will calculate the number of moles of each reactant, the reactant with less number of moles is the limiting reactant.
CO2:
n = m/M where m is the mass and M is the molar mass
n = 40.0g/44.01 g/mol
n = 0.909 mol
C:
n = m/M
n = 15.0 g/12,0107 g/mol
n = 1.249 mol
CO2 is the limiting reagent and C is the reagent in excess.
Grams of CO that will be produced:
The molar ratio between CO2 and CO is 1:2
Therefore the number of moles of CO = 0.909 x 2 = 1.818 mol
m = n x M
m = 1.818 mol x 28,01 g/mol
m = 50.916 g
(c) To find how much of the reagent in excess will be left we will use the stoichiometry
n = 0.909 mol
m = 0.909 mol x 12.0107 g/mol
m = 10.916 g
15.0 g - 10.916 g = 4.084 g
Therefore the mass excess left = 4.084 g