Answer:
- The first picture attached is the diagram that accompanies the question.
- The<u> second picture attached</u> is the diagram with the answer.
Explanation:
In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.
The dissociaton equation for NaCl(aq) is:
- NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)
The dissociation equation for CaCl₂ (aq) is:
- CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.
Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).
Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.
Inorganic molecules are composed of other elements. They can contain hydrogen or carbon, but if they have both, they are organic.
Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
Answer:
An ion with 11 protons, 11 neutrons, and 10 electrons would have a charge of 1+, also expressed as a charge of positive one or +1.