Answer:
The correct answer is - 3.012×10^23 molecules
Explanation:
The number of molecules is present in any mole of an element or compound are the same. To find the molecules you need to multiply the number of the moles of the compound by Avogadro's number:
Moles are calculated from the molecular weight, for AlF3
= 26.98 + 3×19
= 26.98 + 57
= 83.98
Then moles in 42 grams:
= 42/83.98
= 0.5001190 moles
Now the number of molecules = moles*Avogadro's number
= 0.500110×6.022×10^23
= 3.011×10^23 molecules
Answer:
Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)
Explanation:
Let's consider the following molecular equation.
Mg(s) + SnSO₄(aq) ⇄ MgSO₄(aq) + Sn(s)
The full ionic equation includes al the ions and the species that do not dissociate in water.
Mg(s) + Sn²⁺(aq) + SO₄²⁻(aq) ⇄ Mg²⁺(aq) + SO₄²⁻(aq) + Sn(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the species that do not dissociate in water.
Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)
The volume of the unit cell is 2.67 x 10⁻²⁸ m³.
<h3>What is the volume of a unit cell of a body-centered cubic crystal?</h3>
In a body-centered cubic unit cell, the volume occupied by the particles of the substance is about 68% of the total unit cell.
Assuming that a single atomic a sphere, the volume is:
Volume(atom) = 4/3 x π x r³
Volume(atom) = 4/3 x π x (169 x 10⁻¹²)³
Volume(atom) = 2.02 x 10⁻²⁹ m³
There are a total of 9 atoms in a body-centered unit cell, so the total volume occupied by atoms is:
2.02 x 10⁻²⁹ x 9
= 1.82 x 10⁻²⁸ m³
Volume of cell = (1.15 x 10⁻²⁸ ) / 0.68
Volume of cell = 2.67 x 10⁻²⁸ m³
Therefore, the volume of the unit cell is 2.67 x 10⁻²⁸ m³.
Learn more volume of unit cells at: brainly.com/question/1594030
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Kind of this is probably wrong Mass weight of solid, volume weight of liquid?
I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture.
Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
<em>Q = +6970 kJ</em>
