Antimony, Sb, has atomic number 51.
a)WHICH PERIOD AND BLOCK IS ANTIMONY FOUND?
is it period 6 block 13???
b)DIFFERENT ISOTOPES OF ANTIMONY HAVE THE SAME CHEMICAL PROPERTIES. EXPLAIN WHY (1)
Is it their the same thing??
C)A SAMPLE OF ANTIMONY, Ar=121.8, WAS ANALYSED AND WAS FOUND TO CONSIST OF 60% OF 121Sb AND ONE OTHER ISOTOPE. DETERMINE THE MASS NUMBER OF THE OTHER ISOTOPE IN THE SAMPLE OF ANTIMONY.
please help I'm really stuck,
a. no.
b. Because the electron configuration is the same for both isotopes.
c. How did Ar= 121.8 get in there?
If you will go up to the Br isotope problem you posted, just follow what I did for that and you can do this one.
DrBob222 please help with a i dont know what to do or what it is and for c that was what the question said and I'm still confused please tell me the steps specific to this.
IM VERY STUCK ON C
My teacher put in Ar=121.8 so we would know the ar to use if for the question if that helps yuo answer the question so you can hopefully help me
a) To determine the period and block of an element on the periodic table, you need to consider its atomic number. In the case of antimony (Sb), with atomic number 51, it is located in period 5 and block p.
b) Isotopes of an element have the same chemical properties because they have the same number of protons and electrons, which determine the element's chemical behavior. The number of neutrons may vary, thus creating different isotopes. However, since chemical properties are mainly determined by the electron configuration and the number of protons, isotopes of an element exhibit similar chemical properties.
c) To determine the mass number of the other isotope in the sample of antimony, we need to use the given information that 60% of the sample consists of 121Sb isotope.
Let's assume the other isotope is labeled as xSb, where x is the unknown mass number. Since the percentage composition of both isotopes needs to add up to 100%, we can set up the equation:
Percentage of 121Sb + Percentage of xSb = 100%
Given that 60% of the sample is 121Sb, we can substitute the values into the equation:
60% + Percentage of xSb = 100%
To solve for the percentage of xSb, we isolate it on one side of the equation:
Percentage of xSb = 100% - 60%
Percentage of xSb = 40%
This means that the other isotope, xSb, makes up 40% of the sample. Since the atomic mass of antimony (Ar) is 121.8, and given that the 121Sb isotope is already accounted for with a mass number of 121, we can calculate the mass number of the other isotope (xSb):
(40% of Ar - (40% of 121)) / (molar mass of xSb)
Simplifying the equation:
(0.4 * 121.8 - 0.4 * 121) / (molar mass of xSb)
= (48.72 - 48.4) / (molar mass of xSb)
= 0.32 / (molar mass of xSb)
Unfortunately, without the molar mass value for xSb, we can't determine the exact mass number of the other isotope in the sample of antimony.