Answer:
1s22s22p6<u>3s23p4</u>
Explanation:
Sulfur is located in the p block and has 6 valence electrons (the 2 exponent on the 3s and the 4 exponent on the 3p add up to 6)
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
<span> 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
from the reaction 2 mol 4 mol
from the problem 5.4 mol 10.8 mol
M(CO2) = 12.0 +2*16.0 = 44.0 g/mol
10.8 mol CO2 * 44.0 g CO2/1 mol CO2 = 475.2 g CO2 </span>≈480 = 4.8 * 10² g
Answer is C. 4.8*10² g.
Answer:
positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).
