Half reaction: Cr₂O₇²⁻(aq) + 7H₂O(l) + 6e⁻ → 2Cr³⁺(aq) + 14OH⁻(aq).
Chromium change oxidation number from +6 in dichromate ion Cr₂O₇²⁻ to +3 in chromium cation.
Sum of the charges on the left side of the half reaction is -2 and on the left side is -8 (2·(+3) +14·(-1)), so six electrons must be added on the left side of half reaction.
Answer:
The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)
The number of moles = 5 g / 132.14 g/mol = 0.038 mol
The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23
the number of positive ions present in the ammonium sulphate solution:
2 positive ions for every 1 molecule of (NH₄)₂SO₄
so 2 x 2.29x10^23 = 4.58x10^23
the number of negative ions present in the ammonium sulphate solution
1 negative ion for every 1 molecule of (NH₄)₂SO₄
so 1 x 2.29x10^23 = 2.29x10^23
the total number of ions present in the ammonium sulphate solution
4.58x10^23 + 2.29x10^23 = 6.87x10^23
The formula is SrCl2. hope this helps
Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
