Answer is: mass of <span>potassium bromide is 4.71 grams.
V(KBr) = 25.4 mL </span>÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L 0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.
Fe+CuSO4⟶Cu+FeSO4
Given that
FeSO4 = 92.50 g
Number of moles = amount in g / molar mass
=92.50 g / 151.908 g/mol
=0.609 moles FeSO4
Now calculate the moles of CuSO4 as follows:
0.609 moles FeSO4 * 1 mole CuSO4 /1 mole FeSO4
= 0.609 moles CuSO4
Amount in g = number of moles * molar mass
= 0.609 moles CuSO4 * 159.609 g/mol
= 97.19 g CuSO4
<span>Use the Ideal law Equation :
P.V= n.R.T
V = 0.5 L
P = 1.0 atm
</span><span>R= 0.0821 L*atm/mol*K
</span>
<span>n = R*T/P*V
</span><span>P*V= n*R*T
</span>
1.0 * 0.5 = n *<span>0.0821*298
0,5 = n* 24.4658
n = 0,5 / 24.4658
n =0.0204 moles
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