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A certain covalent compound contains sulfur bonded to chlorine. Which of the following statements about this compound is correct?
Question 5 options:
A)
Electrons will reside closer to sulfur, and the bond will be non-polar.
B)
Electrons will reside closer to sulfur, and the bond will be polar.
C)
Electrons will reside closer to chlorine, and the bond will be polar.
D)
Electrons will reside closer to chlorine, and the bond will be non-polar.
Answer:
C) Electrons will reside closer to chlorine, and the bond will be polar
Explanation:
Electrons tend to be closer or attracted to high electronegativity and sulfur has less electronegativity than chlorine. Therefore, electrons will reside closer to chlorine. To find out if it is polar or nonpolar, according to the electronegativity chart you can find online, you can see that the electronegativity of sulfur is 2.5 while the electronegativity of chlorine is 3.0. Because the difference between these two is a slight change, it would be slightly polar. So, the bond will be polar.
Answer:
Electrons will reside closer to chlorine, and the bond will be polar.
Explanation:
ehat is the name of the hydrate Ba(OH)2 8 H2O
Answer: Barium hydroxide octahydrate
Explanation:
If a sample of nitroglycerin containing 2 mL (density = 1.592g/mL) is detonated, how many total moles of gas are produced? If each mole of gas occupies 55L under the conditions of the explosion, how many liters of gas are produced? How many grams of nitrogen gas are specifically produced?
Total moles of gas = 0.1225
Volume of gas produced : 6.7375 L
mass of Nitrogen : 0.588 g
Further explanationGiven
2 ml of Nitroglycerin(ρ=1.592 g/ml)
Required
Total moles of gas
Solution
Nitroglycerin detonated ⇒ decomposition reaction
4C₃H₅N₃O₉(s)⇒ 6N₂(g)+12CO(g)+10H₂O(g)+7O₂(g)
mass of Nitroglycerin :
[tex]\tt mass=2~ml\times 1.592~g/ml=3.184~g[/tex]
moles of Nitroglycerin :
[tex]\tt moles=\dfrac{3.184}{227,0865~g/mol}=0.014[/tex]
Total moles of gas:
[tex]\tt \dfrac{6+12+10+7}{4}\times 0.014=0.1225[/tex]
Volume of gas produced :
[tex]\tt 0.1225\times 55=6.7375~L[/tex]
moles of Nitrogen :
[tex]\tt \dfrac{6}{4}\times 0.014=0.021[/tex]
mass of Nitrogen :
[tex]\tt 0.021\times 28=0.588~g[/tex]