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REY [17]
2 years ago
7

Describe the properties of alkali metals. based on their electronic arrangement, explain whether they exist alone in nature.

Chemistry
1 answer:
saul85 [17]2 years ago
5 0

The alkali metals can't exist alone in nature because of incomplete outermost shell of alkali metals.

<h3>What are the  properties of alkali metals?</h3>

The alkali metals have the high thermal and electrical conductivity. It has high lustre, ductility, and malleability as compared to other materials. Each alkali metal atom has one electron in its outermost shell which make more reactive.

So we can conclude that the alkali metals can't exist alone in nature because of incomplete outermost shell of alkali metals.

Learn more about metal here: brainly.com/question/25597694

#SPJ1

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marshall27 [118]

Mass of sodium thiosulfate Na_{2}S_{2}O_{3} is 110. g

Volume of the solution is 350. mL

Calculating the moles of sodium thiosulfate:

110. g Na_{2}S_{2}O_{3} * \frac{1 mol Na_{2}S_{2}O_{3}}{158.1 g Na_{2}S_{2}O_{3}} = 0.696 molNa_{2}S_{2}O_{3}

Converting the volume of solution to L:

350. mL * \frac{1 L}{1000 mL} = 0.350 L

Finding out the concentration of solution in molarity:

\frac{0.696 mol}{0.350 L} =  1.99 mol/L

4 0
3 years ago
Using Models to Answer Questions About Systems
viva [34]
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5 0
3 years ago
Read 2 more answers
For the following electrochemical reaction: Al3+(aq) + 3e -&gt; Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -&gt; 2F (aq) Calculate
makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=2.87-(-1.66)=4.53V

Hence, the standard electrode potential of the cell is 4.53 V.

6 0
3 years ago
You have 100 g of Pb(NO3)2. How many moles is this?​
topjm [15]

Answer:

0.302 moles

Explanation:

Data given

Mass of Pb(NO₃)₂ = 100 g

Moles of Pb(NO₃)₂ = ?

Solution:

To find mole we have to know about molar mass of Pb(NO₃)₂  

So,

Molar mass of Pb(NO₃)₂  = 207 + 2[14 + 3(16)]

                                           = 207 + 2[14 + 48]

                                           = 207 + 124

Molar mass of Pb(NO₃)₂  = 331 g/mol

Formula used :

                no. of moles = mass in grams / molar mass

Put values in above formula

                   no. of moles = 100 g / 331 g/mol

                   no. of moles = 0.302 moles

no. of moles of  Pb(NO₃)₂ = 0.302 moles

7 0
3 years ago
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Answer:

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Explanation:

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