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svetoff [14.1K]
3 years ago
12

Why are Alkali Metals soft but still extremely reactive?

Chemistry
1 answer:
MrRa [10]3 years ago
8 0
Alkali metals are very reactive. They only have one electron in their outer shell. Because of this they are ready for ionic bonding with other elements to lose that one electron.
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What is the name of the isotope produced when Californium-251 emits an alpha particle?
Charra [1.4K]
<h3>Answer:</h3>

                 Curium-247 <em>i.e.</em> ²⁴⁷₉₆Cm

<h3>Explanation:</h3>

Alpha decay is given by following general equation,

                              ᵃₓA    →    ⁴₂He  +  ᵃ⁻⁴ₓ₋₂B

Where;

           A  =  Parent Isotope

           B  =  Daughter Isotope

           ᵃ  =  Mass Number

           ₓ  =  Atomic Number

Californium-251 is the parent isotope in our case and it has 98 protons (atomic number) and is given as,

                                                 ²⁵¹₉₈Cf

The alpha decay reaction of Californium-251 will be as,

                               ²⁵¹₉₈Cf     →     ⁴₂He  +  ²⁴⁷₉₆B

The symbol for B with atomic number 96 was found to be the atom of Curium (Cm) by inspecting periodic table. Hence, the final equation is as follow,

                               ²⁵¹₉₈Cf     →     ⁴₂He  +  ²⁴⁷₉₆Cm

3 0
3 years ago
What grade percentage is developing, proficient, exceeding, and emerging?​
user100 [1]

Answer:

Grade A is the best percentage that is developing, proficient, exceeding, and emerging

5 0
3 years ago
Read 2 more answers
Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu
nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^0_{[Pb^{2+}/Pb]}=-0.13V

E^0_{[Cu^{2+}/Cu]}=+0.34V

E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0
3 years ago
5. Mr. Martin's Spanish class is 45 minutes long. If it starts at 3:30, what time does it end?
zubka84 [21]
It would be 4:15 . Just add 3:30 plus 45 too get the answer
5 0
3 years ago
Read 2 more answers
What the correct answer?
alexandr402 [8]

B. Bleach and sea water should be identified as bases

Explanation:

  • The pH scale measures the acidic or basic nature of a substance.
  • Ranges from 0 to 14.
  • If pH is lower than 7 → the solution is an acid.
  • If pH is more than 7 → the solution is an basic or alkaline.
  • If a pH is a 7 it is neutral.

In the given question, the pH scale measures for bleach is 8 and for sea water it is 13. So, bleach is basic, not neutral and Sea water is basic too instead of acid. So, Bleach and sea water should be identified as bases.

3 0
3 years ago
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