Answer:
A. m C5H12 = 108.23 g
B. m F2 = 547.142 g
C. m Ca(CN)2 = 71.85 g
Explanation:
- mass (m) = mol (n) × molecular weigth (Mw)
∴ Mw C5H12 = ((12.011)(5)) + ((1.008)(12)) = 72.151 g/mol C5H12
∴ Mw F2 = (18.998)(2) = 37.996 g/mol F2
∴ Mw = Ca(CN)2 = 40.078+((12.011+14.007)(2)) = 92.114 g/mol Ca(CN)2
A. m C5H12 = ( 1.50 mol)×(72.151 g/mol) = 108.23 g C5H12
B. m F2 = (14.4 mol)×(37.996 g/mol) = 547.142 g F2
C. m Ca(CN)2 = (0.780 mol)×(92.114 g/mol) = 71.85 g Ca(CN)2
Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L
Answer:
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Explanation:
This is ur answer.....
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