Answer:
0.47 M
Explanation:
The concentration of the solution can be calculated using the following equation:

<u>Where:</u>
V: is the volume of the solution = 68.6x10⁻² L
η: is the moles of cobalt (II) sulfate
m: is the mass of cobalt (II) sulfate = 89.94 g
M: is the molar mass of cobalt (II) sulfate = 281.103 g/mol
The concentration of cobalt (II) sulfate is:
We used the molar mass of the cobalt (II) sulfate heptahydrate (281.103 g/mol) since it is one of the most common salts of cobalt.
Therefore, the concentration of a solution of cobalt (II) sulfate is 0.47 M (assuming that the cobalt (II) sulfate is heptahydrate).
I hope it helps you!
Answer:
H+ and OCl-
Since this is an acid, the H+ will come off and what is left OCl- is the other part.
Answer:
C. ribosomes
Explanation:
Both prokaryotic and eukaryotic cells contain certain structures called ORGANELLES. They possess some in common and others are not found in one or the other. According to this question, a small, free-floating organelle made from nucleic acid and amino acid is found in both eukaryotes and prokaryotes. This organelle is RIBOSOMES.
Ribosomes are organelles responsible for the synthesis of protein in both eukaryotes and prokaryotes. They can be found free-floating or attached to endoplasmic reticulum. Ribosomes are predominantly made of RNA (nucleic acid) and proteins i.e. Ribosomal RNA (rRNA) and proteins is their structural constituent. Hence, the organelle in this question is RIBOSOME.
We may use the molecular formula of the compound to determine the number of oxygen atoms in one formula unit. The formula unit is:
Mg(NO₃)₂
Here, we can see that there are two nitrate ions in each mole, and each mole of nitrate ion contains three oxygen atoms. Thus, there are 6 oxygen atoms in each formula unit of magnesium nitrate.
By stoichiometry and assume
that:
CxH2xOy + zO2 -> xCO2
+ xH2O
<span>
CO2: 9.48/44 = 0.215 mmol
H2O: 3.87/18 = 0.215 mmol
mass of C = 0.215 * 12 = 2.58 mg
mass of H = 0.215 * 2 * 1 = 0.43 mg
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg
So C/O = 2.58/1.11 ≈ 3 </span>
<span>
Thus we have C3H6O</span>
<span> </span>