The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is 6.18 * 10⁴ Kg of ammonia.
<h3>What mass in kilograms of ammonia are required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄?</h3>
The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is determined from the mole ratio of the reaction.
The mole ratio of the reaction is obtained from the balanced equation of the reaction given below:
- 2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)
Mole ratio of NH₃ and (NH₄)₂SO₄ is 2: 1
Mass of 2 moles of ammonia = 2 * 17 = 34 g
Mass of 1 mole of (NH₄)₂SO₄ = 132 g
Mass of ammonia required = 34/132 * 2.40 × 10⁵ kg
Mass of ammonia required = 6.18 * 10⁴ Kg of ammonia.
In conclusion, the mole ratio is used to determine the mass of ammonia required.
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A) CaCl2
Calcium chloride is an inorganic compound, a salt with the chemical formula CaCl2. It is a colorless crystalline solid at room temperature, highly soluble in water.
<span>E = hxf, où h = constante de Planck = 6,626 x 10 ^ -34Js
E = 6,626 x 10 ^ -34 x 8.11 x 10 ^ 14 = 5.373 10^ -19 J </span><span>
Hope this answers your question, Kimmyers14!</span>
Answer:
Major Product = 4-chloro-4-methylcyclohex-1-ene
Explanation:
Alkene are the class of organic compounds which contain one or more double bonds between two carbon atoms. Alkenes are considered most reactive among the unsaturated hydrocarbons and they undergo <em>addition reactions</em> due to high electron density around the double bonds.
In given question it is written that we are provided with one equivalent of HCl while, our compound contains two double bonds (diene) so in selected starting material the HCl will be added across (hydrohalogenation reaction) the substituted double bond because it will give a more stable carbocation (<u><em>tertiary carbocation</em></u>) during the reaction course. Hence, as shown in reaction scheme 4-chloro-4-methylcyclohex-1-ene will be the major product.
