Answer:
The answer will be 2.98K
Explanation:
Using the formula:
Q = mc∆T
Q= 5,800 (heat in joules)
m= convert 15.2kg to g which is 15200g (mass in grams)
c= 0.128 J/g °c (Specific heat capacity)
∆T= what we need to find (temperature change)
5800J = 15200g x 0.128 x ∆T
= 2.98K
Explanations:- Part 1: We could count the total number of electrons by looking at the electron configurations. Both of these electrons configurations have 47 electrons. If we look at the periodic table then 47 is the atomic number of silver. So, the name of the element is silver and its represented as Ag.
Part 2: As per the rule, Completely filled and half filled orbitals are more stable. First electron configuration has 9 electrons in 4d and we know that d is more stable if it has 5 electrons(half filled) or it has 10 electrons(full filled).
For stability reasons, one of the electron from 5s goes to 4d and for this reason the second electron configuration is found most often in nature for silver.
Few other examples are Cr and Cu.
If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first
<h3>What is EDTA ?</h3>
EDTA is a type of chemical which binds certain metal ions such as calcium and magnesium. some of the functions of EDTA includes:
- Preventing blood clotting of blood samples
- prevention of the formation of Biofilm by bacterias
The EDTA will readily react with metals which have a hiogher number of free ions that it can bind with.
Hence we can conclude that If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first.
Learn more about EDTA : brainly.com/question/10818175
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
According to law of conservation of mass.
In a chemical reaction Mass is neither created nor destroyed.
- Mass of product =Mass of reactant
#1
Mass of water
#2
Mass of sulphur:-
