A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r
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Answer:
The correct answer is option C.
Explanation:
A. The -COOH group would experience hydrogen bonding with water.
The statement is true because , double bonded oxygen atom of carboxylic group interacts with hydrogen atom of water molecule and oxygen atom of water molecule is in same interaction with hydrogen atom of carboxylic acid.
B. The long alkane ''tail" , is hydrophobic.
The statement is true because , region which do not interacts with water molecule is termed as hydrophobic region and region which do interacts with water molecule is termed as hydrophilic region. Generally ,alkanes are are hydrophobic in nature so is the alkane chain in stearic acid.
C. Stearic acid and water are probably miscible.
The statement is false because of higher hydrophobic region the interaction between the acid molecules and water molecules results in immiscibility.
D. Stearic acid would probably dissolve in non polar solvents such as hexane .
The statement is true because of hydrophobic part in stearic acid is larger so that is why it get easily dissolve in non polar solvents.
E. Stearic acid contains hydrophobic regions
The statement is true because of higher hydrophobic region due to which they easily get dissolved in organic and non polar solvents.
Answer:
1. Stąd empiryczny wzór substancji to N₂O₅
2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.
Explanation:
1. Mamy tutaj;
Masa molowa tlenku azotu = 108u
Masa azotu = 14,0067u
Masa tlenu = 15,999 u
74,07% masy tlenku azotu to tlen
Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u
Masa obecnego azotu = 108 - 79,9956 = 28,0044u
Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5
Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2
Stąd empiryczny wzór substancji to N₂O₅.
2. Kiedy sód reaguje z chlorem, mamy;
2Na (s) + Cl₂ (g) → 2NaCl (s)
Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl
W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl
Masa Na obecnego w reakcji = 45 g
Masa molowa sodu = 22,989769u
Liczbę moli sodu w 45 g sodu podano w następujący sposób;
Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl
Masa molowa NaCl = 58,44 g / mol
Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl
Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g
W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.