Answer: The noble gas preceding tin is Krypton [Kr].
Explanation: Atomic Number of tin is 50 , Atomic Number of neon is 10, Atomic Number of argon is 18, Atomic Number of krypton is 36 and Atomic Number of xenon is 54.
Atomic number tells us the number of electrons present in the atom.
Thus the electronic configuration of tin according to increasing energy of orbitals is written as ![[Kr]4d^105s^25p^2](https://tex.z-dn.net/?f=%5BKr%5D4d%5E105s%5E25p%5E2)
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Answer:
37.5 g NaCl
Explanation:
Step 1: Given data
- Concentration of NaCl: 15.0% m/m
- Mass of the solution: 250.0 g
Step 2: Calculate how many grams of NaCl are in 250.0 g of solution
The concentration of NaCl is 15.0% by mass, that is, there are 15.0 g of NaCl every 100 g of solution.
250.0 g Solution × 15.0 g NaCl/100 g Solution = 37.5 g NaCl
Because the easier a metal donates an electron (Low Ionization Energy), the more readily a non-metal with high electromagnetically will accept it.
Thus they match well, - one readily donates, the other readily accepts. This is how ionic bonding occurs.
Hope this helps.
The number of protons uniquely identifies the element. This is not on the periodic table. Isotopic notationincludes the atomic number and the mass number. 8 17 This isotope of oxygen has 8 protons (atomic number), 8 electrons and 9 neutrons
Answer:
The answers to your question are below
Explanation:
a) 6.85×1020 H2O2 molecules
H2O2 MW = 32 + 2 = 34 g
34g -------------------- 6.023 x 10²³ molecules
x ------------------- 6.85 x 10 ²⁰
x = (6.85 x 10 ²⁰)(34)/ 6.023 x 10²³
x = 0.038 g
3.3×1022 SO2 molecules
MW SO2 = 32 + 32 = 64g
64 g -------------------- 6.023 x 10²³ molecules
x -------------------- 3.3×1022 SO2 molecules
x = (3.3×1022 SO2)(64) / 6.023 x 10²³
x = 3.51 g
5.5×1025 O3 molecules
MW = 16 x 3 = 48g
48 g ----------------- 6.023 x 10²³ molecules
x ------------------ 5.5×1025 O3 molecules
x = (5.5×1025 )(48) / 6.023 x 10²³
x = 4383 g
9.30×1019 CH4 molecules
MW = 12 + 4 = 16 g
16 g -------------------- 6.023 x 10²³ molecules
x -------------------- 9.30×1019 CH4 molecules
x = (9.30×1019)(16) / 6.023 x 10²³
x = 0.0025 g
