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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You did not include the options but I can tell you the product ratio.
The product ratio is the mole ratio of the products of the reaction.
From the balanced chemical equation you have all the mole ratios:
The given equation is: 2 C6H5COOH + 15O2 --> 14 CO2 + 6H2O
The mole ratios are: 2 C6H5COOH: 15 O2: 14 CO2 : 6 H2O
The products are CO2 and H2O
Their mole ratio = 14 CO2 : 6 H2O
That can be expressed as:
14 mol CO2 7 mol CO2
----------------- = -----------------
6 mol H2O 3 mol H2O
It is also the same that:
6 mol H2O : 14 mol CO2
6 mol H2O 3 mol H2O
------------------ = -------------------
14 mol CO2 7 mol CO2
So, compare your options to the ratios show above and pick the proper ratio.
Answer:
Ions and ionic bonds. Some atoms become more stable by gaining or losing an entire electron (or several electrons). When they do so, atoms form ions, or charged particles. Electron gain or loss can give an atom a filled outermost electron shell and make it energetically more
Answer:
1. 0.00352 M
2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3. 0.00534 M
Explanation:
1.
Mass of strontium hydroxide= 10.45 g
Volume of solution = 41.00 ml
Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles
Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M
2.
2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3.
Concentration of acid CA= the unknown
Volume of acid VA= 31.5 ml
Concentration of base CB= 0.00352 M
Volume of base VB= 23.9 ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB = NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
CA= 0.00352 × 23.9 ×2/31.5 ×1
CA= 0.00534 M
