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Kobotan [32]
3 years ago
14

Which metal in Period 5 is very reactive and has two valence electrons in each atom? rubidium (Rb) strontium (Sr) zirconium (Zr)

silver (Ag)
Chemistry
2 answers:
Paul [167]3 years ago
6 0

Answer:

strontium (Sr)

Explanation:

jus took the quiz

vladimir1956 [14]3 years ago
4 0
<h3>Answer:</h3>

           Strontium (Sr)

<h3>Explanation:</h3>

                        The condition given in statement is the presence of two valence electron. Hence, first we found the electronic configuration of given atoms as follow;

                       Rubidium                         [Kr] 5s¹

                      Strontium                        [Kr] 5s²

                       Zirconium                         [Kr] 4d² 5s²

                      Silver                                 [Kr] 4d¹⁰ 5s¹

From above configurations it is cleared that only Strontium and Zirconium has two electrons in its valence shell.

We also know that s-block elements are more reactive than transition elements due to less shielding effect in transition elements hence, making it difficult for transition metals to loose electrons as compared to s-block elements. Therefore, we can conclude that Strontium present in s-block with two valence electrons is the correct answer.

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The symbol of the period five element that is a member of the pnicitides family are antimony.

Explanation:

Pnictogen family

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  • There are five valence electrons each member of pnictogen family. In group 15 double bonds and triple bonds are formed due to these valence electrons.  
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Antimony  

  • Antimony is the element found in period 5 and block p.
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7 0
3 years ago
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume
MissTica

Answer:

4.56 X 10^ -4 g/mL

Explanation:

A solution is  prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.

(7.6 X10^-4  gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g

this is dissolved )in 10 m L=45.6 X 10^-4  g/ 10

4.56 X 10^ -4 g/mL

check

6/10 =0.6

4.56/7.6 = o.,6

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Answer:

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Explanation:

Hello there!

In this case, since the combustion of butane is:

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Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}}  *\frac{18.02gH_2O}{1molH_2O}

Therefore, the resulting mass of water is:

m_{H_2O}=11.3gH_2O

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