The molar mass of Na₂SO₄ -
2 x Na - 2 x23 = 46
1 x S - 1 x 32 = 32
4 x O - 4 x 16 = 64
total = 46 + 32 + 64 = 142 g/mol
the molarity of solution - 2.0 M
in 1 L of solution , 2.0 moles
Therefore in 2.5 L - 2 mol/L x 2.5 L = 5 mol
then the mass of Na₂SO₄ required = 142 g/mol x 5 mol = 710 g
We are going to use this equation:
ΔT = - i m Kf
when m is the molality of a solution
i = 2
and ΔT is the change in melting point = T2- 0 °C
and Kf is cryoscopic constant = 1.86C/m
now we need to calculate the molality so we have to get the moles of NaCl first:
moles of NaCl = mass / molar mass
= 3.5 g / 58.44
= 0.0599 moles
when the density of water = 1 g / mL and the volume =230 L
∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg
now we can get the molality = moles NaCl / Kg water
=0.0599moles/0.23Kg
= 0.26 m
∴T2-0 = - 2 * 0.26 *1.86
∴T2 = -0.967 °C
Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.
Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."
The atom has only one isotope which means 100 % of same atom is present in nature. The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. Mass of one atom of that isotope is 9.123 ✕ 10⁻²³ g, so mass of one mole of atom that is Avogadro's number of atom is 6.023 X 10²³ X 9.123 X 10⁻²³ g=54.94 g = 55 g (approximate).
So, the atom having atomic mass 55 will be Cesium (Cs). Only one isotope of Cesium is stable in nature.
Answer:
Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>
- It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
- Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
- The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.
- <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>
∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.
- From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
- So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
- Finally, we can calculate the mass of copper produced using:
mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.
- <u><em>So, the answer is:</em></u>
<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>
