Answer:
K will give up an electron more easily than Br.
Explanation:
Electronegativity of an element is a property that combines the ability of its atom to lose and gain electrons.
The lower the electronegativity value, the more electropositive an element is and the more readily it loses electrons.
From the data given, we see that Br has an E.N value of 3.0 and K has an E.N value of 0.82.
Therefore, Br is highly electronegative and it is able to attract electrons to itself whereas K has a low E.N value. K will give up electrons more readily.
Lookinf at other information in the table, the larger atomic radius and lower ionizaton energy of K are all pointers to how readily it would be able to lose electrons.
We can conclude that K is even a metal.
<span>I believe its, the federal principle or system of government. Hope this can help!! >-<</span>
Page 1
1: The grass
2: The grasshopper
3: The frog
4: The snake
5: The eagle
Page: 2
The answer is the 2nd one
Page: 3
The owl would because its at the end it would have almost no more sun energy left in whatever it eats.
Hope This Helps :)
Answer:
The answer to your question is: The mass number will be 4 units lower.
Explanation:
Alpha particles are Helium atoms, which have a mass number of 4 and atomic number of 2.
When an alpha particle is released, the original atom loses 2 protons and and 2 neutrons an we can see in the example.
²²⁶ ₈₈ Ra ⇒ ²²² ₈₆ Rn + ⁴₂ He
Hey there!:
Given the mass of PbCl(OH) :
0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g
Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol
Atomic mass of Pb = 207 g/mol
Hence mass of Pb in 135 g PbCl(OH) :
(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =
0.79768 * 135 => 107.68 g of Pb
For Pb2Cl2CO3 :
Given the mass of Pb2Cl2CO3 :
0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g
Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol
Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g
Hence mass of Pb in 135 g Pb2Cl2CO3:
(414 g Pb / 545 g PbClOH) * 135g PbClOH =
0.75963 * 135 => 102.55 g of Pb2Cl2CO3
Hope that helps!
