Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
Spontaneous = exothermic. If it is not spontaneous, it is endothermic. ΔG > 0, which means that the reaction uses energy instead of emitting it.
The phenomenon that naturally warms the earth's lower atmosphere and surface is called the greenhouse effect.
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Answer:
The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.
Explanation:
The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.
The parameters of this equation are:
P = 1.918 atm
V = 750.0mL = 0.75L
n = ?
R = 0.0821
T = 25 degree celcius = 25 + 273 = 298 degree kelvin.
From this formular, n = (PV)/(RT)
n = (1.918 X 0.75)/(0.0821 X 298 )
n = 0.0588
n, no of mole = mass/molar mass
0.0588 = 8.6/MM
MM = 8.6/0.0588
MM = 146.25g/mole.
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