The answer is there is a one-to-one ratio of potassium ions to iodide ions.
Explanation :
- (K) belongs to Alkali metals in group (1A) that contains (1) electron in the outermost energy level, whereas, (I) is from halogens in group (7A) that contains (7) electron in the outermost energy level.
- To achieve stability, both atoms tend to reach the nearest noble state (outermost level occupies 8 electrons). Therefore, (K) loses its outer electron and gives it to (I) which now has a completely filled outer level and an ionic bond is formed between the two.
- The valency (number of electrons lost, gained or shared) of both atoms is equal ”monovalent” which means one-to-one ratio..
Answer:
B) -4.1 units
Explanation:
According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:
- first increase by 3.6 units
- then increase by another 18.7 units
- then decrease by 12.2 units
- and finally attains a value of 85.5 units
This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5
To get x, we say;
89.6 + 3.6 = 93.2
93.2 + 18.7 = 111.9
111.9 - 12.2 = 99.7
99.7 - 85.5 = 14.2units.
The changes that occured is represented as follows:
= (3.6 + 18.7) - (12.2 + 14.2)
= 22.3 - 26.4
= -4.1 units
The balanced equation is:
BaCl2 (aq) + Na2SO4 (aq) ----> BaSO4(s)+ 2 NaCl(aq)
This is a double replacement reaction.
The reactants are:
a) BaCl2: barium chloride, a ionic compound, therefore soluble in water,
b) Na2SO4: sodium sulfate, another ionic compound, therefore also soluble in water.
The products are:
c) BaSO4: barium sulfate, a solid not soluble in water which precipitates.
d) NaCl: sodium chloride, an ionic compound, therefore soluble in water.
Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
