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Iteru [2.4K]
3 years ago
6

How many orbitals are available in the first energy level?

Chemistry
2 answers:
velikii [3]3 years ago
5 0

Answer:

1

Explanation:

a p e x :)

Archy [21]3 years ago
4 0
I believe it may be two, my good fellow.
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What is a compound in chemistry
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Two types of chemical bonds common in compounds are covalent bonds and ionic bonds. The elements in any compound are always present in fixed ratios. Example 1: Pure water is a compound made from two elements - hydrogen<span> and </span>oxygen<span>.</span>
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3 years ago
Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden
LenKa [72]

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

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4 years ago
The development of a nuclear power plant at an ocean site is expected to produce an enormous amount of electric power for a larg
Oksanka [162]

Answer:

B. The effects of warmed water on aquatic life

Explanation:

quizizz

8 0
3 years ago
Read 2 more answers
Suppose you are performing a gas-producing reaction with an unknown metal, X . X ( s ) + 2 H C l ( a q ) ⟶ X C l 2 ( a q ) + H 2
maksim [4K]

Answer:

600,000,000,000,000,000

Explanation:

6 0
3 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0
3 years ago
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