Answer: 530 hours
Explanation:
The reduction of Nickel ions to nickel is shown as:
of electricity deposits 1 mole of Nickel
1 mole of Nickel weighs = 58.7 g
Given quantity = 18.0 kg = 18000 g (1kg=1000g)
58.7 g of Nickel is deposited by 193000 C of electricity
18000 g of Nickel is deposited by = 
of electricity
where Q= quantity of electricity in coloumbs = 59182282.8C
I = current in amperes = 31.0 A
t= time in seconds = ?
(1h=3600 sec)
Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A
The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
Answer:
<h2>F=Gm1m2r=G×1×11=G</h2>
Explanation:
<h2>______________________________</h2>
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<h2>
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<span>they assign a numerical date to each rock layer studied.</span>
