Answer : BaS
will be the precipitate which will be formed.
Explanation : When all the three solutions namely;
are mixed together a white precipitate of BaS
is formed as a product in the solution along with the soluble by product of Ammonium nitrate which is
<span>It is known
that acids compounds contains hydrogen and produces hydrogen ion in water. A binary
acid however is an acid that have two elements, one of the element has a
hydrogen attached to it. Examples of binary acids are hydrogen fluoride (HF),
hydrogen bromide (HBr) and hydrogen sulfide (H2S). In naming a binary acid, it
has two rules; one, as pure compounds and two, as acid solutions. For pure
compounds, start with the name ‘hydrogen’ and end the anion name with ‘-ide’. For
acidic compounds, start with ‘hydro-‘, end the anion with ‘-ic’ and add ‘acid’.</span>
Answer:
Identify the "given" information and what the problem is asking you to "find."
Given : Cl2O7
Find: % Composition (% Cl and %O)
List other known quantities.
Mass of Cl in 1 mol Cl2O7 , 2 Cl : 2 x 35.45 g = 70.90 g
Mass of O in 1 mol Cl2O7 , 7 O: 7 x 16.00 g = 112.00 g
Molar mass of Cl2O7 = 182.90 g/mol
Answer:
1) 7.256 mol Br2 (Cl2)/(Br2)
The Br2 cancels out, so we have 7.256(2)
This is 14.512.
2) Number of moles = mass / molar mass
Number of moles = 239.7 g/ 35.5 g/mol
Number of moles = 6.8 mol
BrCl=13.6 mol
13.6(11.5.357)
1568.9 g
3) Repeat the same process with problem 2, given that there are 6.022x10^23 atoms in a mole.
Explanation:
Br2 + Cl2 → 2BrCl
Answer:
15%
Explanation:
From the question given above, the following data were obtained:
Mass of sugar = 15 g
Mass of water = 85 g
Percentage of sugar in the solution =?
Next, we shall determine the mass of the solution. This can be obtained as follow:
Mass of solute (sugar) = 15 g
Mass of solvent (water) = 85 g
Mass of solution =?
Mass of solution = mass of solute + mass of solvent
Mass of solution = 15 + 85
Mass of solution = 100 g
Finally, we shall determine the percentage of the sugar in the solution. This can be obtained as follow:
Mass of solute (sugar) = 15 g
Mass of solution = 100 g
Percentage of sugar in the solution =?
Percentage = solute /solution × 100
Percentage = 15 / 100 × 100
Percentage of sugar in the solution = 15%