Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).
Explanation:
This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7
Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4
<span>7.39 ml
For this problem, simply divide the mass of mercury you have by it's density.
100 g / 13.54 g/ml = 7.3855 ml
Since we only have 3 significant digits in 100., you need to round the result to 3 significant digits. So
7.3855 ml = 7.39 ml</span>
Answer:
1.69.
Explanation:
- The solution = 12.0 / 7.11 = 1.687 = 1.69.
- The rule of significant figures for division states that: the results are reported to the fewest significant figures.
- 12.0 contains 3 significant figures.
- 7.11 contains 3 significant figures.
So, the solution should contain 3 significant figures.
- Now, the issue id of rounding; In a series of calculations, carry the extra digits through to the final result, then round.
- If the digit to be removed is equal to or greater than 5, the preceding digit is increased by 1.
- The digit that should be removed is 7 that is larger than 5 so increase the preceding digit by 1.
Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the 
orbit of hydrogen atom.
Formula used :
where,
= energy of 
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.
The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV
Answer:
Yes Concurred,but where is the question
