Answer:
Hg
Explanation:
We are given information about the unknown element, and using each characteristic, we can narrow down the possible elements until we have just one possibility left.
Conducts electricity: This means that the element has to be a metal, or a semi-metal, because non-metals cannot conduct electricity.
Forming chloride and oxide ions: When it is seen that it forms chloride and oxide ions in the form of XCl₂ and XO, it can be seen that this element has an ionic charge of +2. This narrows it down to the elements in <u>Group 2.</u>
Liquid at room temperature: This is tricky, because we realize that there are no elements in group 2 that are liquid at room temperature. So hence we can look at groups 3 to 12, and see if there are any liquid metals with an ionic charge of +2.
Taking all this information into account, we can see that the only element it can be is mercury (Hg).
Answer:
Do to half of the mnairals this can not be made into a lab there is an error
Explanation:
Answer:
Temperature at which molybdenum becomes superconducting is-272.25°C
Explanation:
Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.
As given, molybdenum becomes superconducting at temperatures below 0.90 K.
Temperature in Kelvins can be converted in °C by relation:
T(°C)=273.15-T(K)
Molybdenum becomes superconducting in degrees Celsius.
T(°C)=273.15-0.90= -272.25 °C
Temperature at which molybdenum becomes superconducting is -272.25 °C
Answer:
4.86×10^23 molecule of Pb
Explanation:
Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.
So:
2 mol NH3/ 3 mol Pb
Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:
(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb
Then, we just need to use Avagadro's number to get the number of molecules.
(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb
Answer:
The atomic mass of second isotope is 7.016
Explanation:
Given data:
Average Atomic mass of lithium = 6.941 amu
Atomic mass of first isotope = 6.015 amu
Relative abundance of first isotope = 7.49%
Abundance of second isotope = ?
Atomic mass of other isotope = ?
Solution:
Total abundance = 100%
100 - 7.49 = 92.51%
percentage abundance of second isotope = 92.51%
Now we will calculate the mass if second isotope.
Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
6.941 = (6.015×7.49)+(x×92.51) /100
6.941 = 45.05235 + (x92.51) / 100
6.941×100 = 45.05235 + (x92.51)
694.1 - 45.05235 = (x92.51)
649.04765 = x
92.51
x = 485.583 /92.51
x = 7.016
The atomic mass of second isotope is 7.016
