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:
The correct answer is C: A radiation consisting of uncharged particles is emitted when alpha particles strike beryllium atoms.
Explanation:
Chadwick discovered neutron while experimenting with a gold foil. A stream of alpha particles produced from a polonium source was directed at beryllium target. It was noticed that some penetrating radiations were produced. These uncharged particles were called neutrons because on the charge detector these particles showed no deflection.
Their nuclear reaction is as follow.
₂He + ₄Be ----------> ₀n + ₆C
Chadwick noticed that neutrons cannot ionize gases and are highly penetrating particles.
Anaerobic transformations of 1,1,1-trichloroethane (TCA), 1,1-dichloroethane (DCA), and chloroethane (CA) were studied with sludge from a lab-scale, municipal wastewater sludge digester. TCA was biologically transformed to DCA and CA and further to ethane by reductive dechlorination. TCA was also converted to acetic acid and 1,1-dichloroethene (11DCE) by cell-free extract. 11DCE was further biologically converted to ethene. This pathway was confirmed by transformation tests of TCA, DCA and CA, by tests with cell-free extract, and by chloride release during TCA degradation.
Diane Julie Abbott is a British Labour Part Politician.
She has been the Member of Parliament for Stoke Newington since 1987.
She became the first black woman to be elected to the House of Commons.
Born in September 27th 1953
Married to David Thompson.
Education: Newnham College, University of Cambridge
Answer:
The final volume of NaOH solution is 30ml
Explanation:
We all know that
V1S1 = V2S2
or V1= V2S2÷S1
or V1= V2×S2×1/S1
or V1=100×0.15×1/0.50
V1= 30
∴30 ml NaOH solution is required to prepare 0.15 M from 100ml 0.50 M NaOH solution.
