Mass of (NH₄)₂U₂O₇ : 410.05 kg
<h3>Further explanation</h3>
Reaction
2UO₂SO₄ + 6NH₃ + 3H₂O → (NH₄)₂U₂O₇ + 2(NH₄)₂SO₄
MW UO₂SO₄ : 366.091
MW (NH₄)₂U₂O₇ : 624.131
MW H₂O : 18.0153
MW NH₃ : 17.0306
mol of 100 kg water :
mol of 100 kg ammonia :
mol of UO₂SO₄ :
Limiting reactants : smallest mol ratio(mol : coefficient)
UO₂SO₄ ⇒ Limiting reactants
mol (NH₄)₂U₂O₇ : mol UO₂SO₄
mass (NH₄)₂U₂O₇
Answer:
See explanation below
Explanation:
To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:
m = m₀e^-kt (1)
In this case, k will be the constant rate of this element. This is calculated using the following expression:
k = ln2/t₁/₂ (2)
Let's calculate the value of k first:
k = ln2/2.7 = 0.2567 d⁻¹
Now, we can use the expression (1) to calculate the remaining mass:
m = 8.1 * e^(-0.2567 * 2.6)
m = 8.1 * e^(-0.6674)
m = 8.1 * 0.51303
m = 4.16 mg remaining
Answer: sorry i’m late but the answer is D- wavelength :)
Explanation:
Answer: 2.8275grams
Explanation: A buffer is made btw a weak acid and it salt. In a solution made by dissolving a weak acid in solution, equilibrium is set up btw ionised and unionised ion. For Benzoic acid
C6H5COOH....> C6H5COO- + H+
Ka = [C6H5COO-] [H+]/ [C6H5COOH].......(1)
using Ka = 6.5× 10^-5, [C6H5COOH] = 0.02M. PH= - log[H+] ....> [H+]= 10^-4M.
Putting the values in(1)
[C6H5COO-]= 6.5× 10^-5 × 0.02/ 10^-4
[C6H5COO-] = 0.013M = Molarity of sodium benzoate
Mole(C6H5COONa) = 0.013 × Volume = 0.013mol/litre × 1.5 litre
Mole(C6H5COONa) = 0.0195mol
Mass(C6H5COONa) = 0.0195 × Molar mass
Mass(C6H5COONa) = 2.8275g
