The percentage adds up to 100%.
<h3>Percentages</h3>
Mass of mixture = mass of container+mixture - mass of container = 56.779 - 54.558 = 2.221 g
CuCO3 is insoluble in water. Thus:
Mass of CuCO3 = mass of beaker and residue - mass of beaker = 78.875 - 77.575 = 1.300 g
Mass of NaCl = mass of mixture - mass of CuCO3 = 2.221 - 1.300 = 0.921 g
%NaCl (w/w) = weight of NaCl/weight of mixture = 10.921/2.221 = 41.468%
% (w/w) CuCO3 = weight of CuCO3/weight of mixture = 1.300/2.221 = 58.532%
Addition of percentages = 41.468 + 58.532 = 100%
More on percentages can be found here: brainly.com/question/24159063
#SPJ1
<h3>
Answer:</h3>
Anion present- Iodide ion (I⁻)
Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)
<h3>
Explanation:</h3>
In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.
Additionally we need to know the color of the precipitates.
Some of insoluble salts of silver and their color include;
- Silver chloride (AgCl) - white color
- Silver bromide (AgBr)- Pale cream color
- Silver Iodide (AgI) - Yellow color
- Silver hydroxide (Ag(OH)- Brown color
With that information we can identify the precipitate of silver formed and identify the anion present in the sample.
- The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
- Therefore, the anion that was present in the sample was iodide ion (I⁻).
- Thus, the corresponding net ionic equation will be;
Ag⁺(aq) + I⁻(aq) → AgI(s)
Answer:
density of a piece of metal = 7 gr/ml
Explanation:
See the file please
Answer:
30 L H2
Explanation:
- 10 L N2 x <u>3 L H2</u> = 30 L H2
. 1 L N2
Try to verify my answer, Stoichiometry is not easy for me.
