Answer:
c. H₂O (l) → H₂O(s)
Explanation:
The correct answer is option C: The molecular formulas for the substance(s) on the left side of the equation and on the right side of the equation are identical. This means that there was not a chemical change, but a physical one.
In this concrete example, what happened was a <em>change of states of matter</em>: from liquid (l) to solid (s).
Other reactions have different molecular formulas on both sides of the equation, thus they represent chemical changes.
Answer:
fluorine (F), and chlorine (Cl) because they are both halogens and belong to the same group in the periodic table
Explanation:
The pair of non - metals in the elements given that will have similar properties are chlorine and Fluorine.
This is because they both are elements commonly known by the group name Halogens which belong to group 7A.
Properties of halogens include;
- They are very reactive, have single covalent bonds and readily form other compounds.
- In the gaseous state, they will form diatomic covalent molecules.
Whereas, helium is a group 8 element, oxygen is a group 6 element while carbon is a group 4 element.
Answer:
0.6 L
Explanation:
As this problem deals with a <em>dilution process</em>, we can solve it by using the following formula:
Where subscript 1 stands for the initial concentration and volume, while 2 stands for the diluted solution's. That means that in this case:
We <u>input the data</u>:
- 18 M * 100 mL = 3.0 M * V₂
And <u>solve for V₂</u>:
Finally we <u>convert 600 mL to L</u>:
Answer:
Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M
≈0.350 M
Explanation:
From the question it is clear that,
Initial volume of sodium sulphide solution is (v₁) = 50mL
Initial concentration of sodium sulphide solution is (s₁) =0.874 M
Final volume of sodium sulphide solution is (v₂) = 250mL
Let, the final concentration of sodium sulphide solution is s₂, then according to acidimetry-alkalimetry,
v₁ × s₁ = v₂ × s₂
Or, s₂ = v₁ × s₁/v₂
= 50 × 0.874 / 250
= 0.1748 M
Therefore, concentration of 250mL sodium sulphide solution is 0.1748 M
Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M
≈0.350 M
Answer: 2.30 grams
Explanation: First find how many moles you have:
The molar masses are taken from the periodic table: Ca is 40, O is 16, H is 1. There are two O and 2 H from (OH)2 40+32+2=74 g/mol. Grams (given) x mol/grams (molar mass) = mols, so 85 grams x 1 mol / 74 grams = 1.149 mol. For every mol of Ca(OH)2, there is 2 mol of H so x2 = 2.297 mol H. For every mol of H, there is one gram so 2.297 grams. Rounded to three significant figures you get 2.30 grams.