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4 years ago

What period are the following elements in He,Ge,Rb,l

Chemistry

1 answer:

patriot [66]4 years ago

5 0

He is in period 1

Ge is in period 4

Rb is in period 5

I is in period 5 as well

-Seth

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6. What is the molarity of 175 mL of solution containing 2.18 grams of NazS04-10H2O?

Ymorist [56]

Answer:

$Molarity\,\,of\,\,the\,\,solution\,\,is\,\,S=0.039M$

Explanation:

$W=2.18 g\\M=322g\\V=175mL=0.175L\\\\S=\frac{W}{MV} \\=>S=\frac{2.18}{322*0.175} \\So,S=0.039M$

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3 years ago

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3 years ago

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What is the polarity of a Na-Br bond?

ivanzaharov [21]

Answer:

likely ionic

Explanation:

The bond between , Sodium and Bromine , in sodium bromide salt , is ionic in nature .

Since ,

The difference in the electronegativity of the two atom is high .

According to Pauling scale ,

The Electronegativity of Bromine = 2.96

The Electronegativity of sodium = 0.93

Hence ,

The Difference in electronegativity = ( 2.96 - 0.93 ) = 2.03

Hence ,

The electronegativity difference value is more than 1.8 ,

Therefore,

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4 years ago

What does unalienable mean? use it in a sentence

Zarrin [17]

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Explanation:

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4 years ago

Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of

dybincka [34]

Answer: The electronic configuration of the elements are written below.

Explanation:

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

Option a: Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is $1s^22s^22p^2$

Option b: Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is $1s^22s^22p^63s^23p^3$

Option c: Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is $1s^22s^22p^63s^23p^64s^23d^3$

Option d: Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3$

Option e: Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6$

Hence, the electronic configuration of the elements are written above.

4 0

3 years ago