Answer:
17.6kg
Explanation:
mass = Volume × Density
mass = 5000cm^3 × 3.52g/cm^3
mass = 17600g = 17.6kg
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.
Learn more about mole ratio at: brainly.com/question/19099163
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Answer:
true
Explanation:
the ocean crust is found at the edges of the ocean soooooooooooo
The Phase of the catalyst is different. (1)
Answer:
79.9 amu
Explanation:
Given data:
Atomic mass of bromine = ?
Percent abundance of 1st isotope = 50.7%
Atomic mass of 1st isotope = 78.92 amu
Percent abundance of 2nd isotope = 49.3%
Atomic mass of 2nd isotope =80.92 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50.7×78.92)+(49.3×80.92) /100
Average atomic mass = 4001.24 + 3989.36 / 100
Average atomic mass = 7990.6 / 100
Average atomic mass = 79.9 amu.
