Answer:
Ksp = 1.07x10⁻²¹
Explanation:
Molar solubility is defined as moles of solute can be dissolved in 1L.
Ksp for NiS is defined as:
NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)
Ksp = [Ni²⁺] [S²⁻]
As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.
Replacing:
Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]
<em>Ksp = 1.07x10⁻²¹</em>
<em></em>
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.
Answer:
36.8 L
Explanation:
We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 80 °C
T(K) = 80 + 273
T(K) = 353 K
Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:
Number of mole (n) = 1.27 moles
Temperature (T) = 353 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 1.27 × 0.0821 × 353
V = 36.8 L
Thus, the volume occupied by the helium gas is 36.8 L
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>
Therefore;
Moles of the compound will be;
= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules
