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ikadub [295]
3 years ago
15

Rfggxchhgfdxvnjgdzvbhdazcbhds

Chemistry
1 answer:
kotegsom [21]3 years ago
5 0

Answer:

hgdvsghsahfabvdhvx

Explanation:

is this a crazy chemistry contest

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A. Balance the following equation:
yanalaym [24]

Answer:

1.C+2H2-CH4

Explanation:

when we add two to hydrogen the product is equal to the reactant

7 0
4 years ago
Each element possesses unique chemical and physical properties. true or false
g100num [7]
True. An element's properties are based on their atomic numbers, atomic masses and electron configurations.


Hope it helped!
8 0
3 years ago
The fertilizer ammonium sulfate, (NH4)2SO4, is prepared by the reaction between ammonia (NH3) and sulfuric acid:
kumpel [21]

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

#SPJ1

4 0
2 years ago
Which of the followings statements accurately describes renewable energy sources?
grigory [225]
Number one is the correct answer
8 0
4 years ago
The radius of a krypton atom is 110pm and it's mass is 1.39x10 to -22 g.
Readme [11.4K]

Answer:

24.9~g/cm^3

Explanation:

Density is found dividing mass by volume. In this case, we treat krypton as a sphere having a volume of V = \dfrac{4}{3}\pi r^3. Given:

m = 1.39\cdot 10^{-22}~g

r = 110 pm = 1.10\cdot 10^{-8}~cm

We obtain density of:

d = \dfrac{m}{V} = \dfrac{m}{\dfrac{4}{3}\pi r^3} = \dfrac{3m}{4\pi r^3}

d = \dfrac{3\cdot 1.39\cdot 10^{-22}~g}{4\pi\cdot (1.10\cdot 10^{-8}~cm)^3} = 24.9~g/cm^3

This is not a feasible value for a gas like krypton, its radius is actually not 110 pm.

6 0
3 years ago
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