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
Mol sulfuric acid = 19 g * (1 mol) / (98.1 g) = 0.19367 mol
mol H2O = 0.19367 mol H2SO4 * (2 H2O) / (1 H2SO4)
= 0.387359 mol H2O
grams H2O = 0.387359 mol H2O * (18 g)/(1 mol)
= 6.97 g
The answer is 7.0 grams of water
Given mass of Scandium = 50.0 g
Increase in temperature of the metal when heated = 
Heat absorbed by Scandium = 
The equation showing the relationship between heat, mass, specific heat and temperature change:
Where Q is heat =
m is mass = 50.0 g
ΔT =
On plugging in the values and solving for C(specific heat) we get,
=50.0g(C)()
C = 0.491
Specific heat of the metal = 0.491
