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Anit [1.1K]
2 years ago
15

Write the expected ground electronic configuratio for the noble gas with electron occupying 4f orbitals?

Chemistry
1 answer:
lisov135 [29]2 years ago
3 0

Answer:

There are main six noble gases in the periodic table. The filling of 4f-orbitals occurs after 6s orbitals. The noble gas that belongs to the sixth period is Radon. Its electronic configuration is [ X e ] 6 s 2 4 f 14 5 d 10 6 p 6  .

Therefore, the name of noble gas is Radon.

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Does a negative exponent mean that the number is less than 1 yes or no​
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Answer:

When a number is written in scientific notation, the exponent tells you if the term is a large or a small number. A positive exponent indicates a large number and a negative exponent indicates a small number that is between 0 and 1.

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2 years ago
What is the kinetic energy of a 5 kg object moving at 7 m/s?
valina [46]

Answer:

122.5 Joule, or 122.5 j

Explanation:

Given,

mass of the object (m) = 5 kg

velocity of the object (v) = 7 m/s

Kinetic energy = \frac{1}{2} × m × v²

Applying the formula:

Kinetic energy = \frac{1}{2} × 5 × 7²

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Kinetic energy is the energy that an object gains as the result of the motion.  It also depends on the mass of the object and force with which the motion is applied.  In the given question, the mass of the object is 5 kg and the force of the velocity by which it is moving is 7 m/s.

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3 years ago
Choose the number of significant figures indicated. 0.06
jeka94
One-
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The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

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