Answer:
Two, KCl and PbCl₂.
Explanation:
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In this case, since the addition of chloride ions promote the following three ionic reactions:

We can infer that both silver chloride and lead (II) chloride are precipitated products as their Ksp are 6.56x10⁻⁴ and 1.59x10⁻⁵ respectively, which means they are merely soluble in water.
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The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
There is 118 elements<span> in the periodic table</span>
Answer:
3
Explanation:
left side has 2 N so right side must have a 2 which means 6 H on right side so to get 6 on left you have a coef. of 3 to make 6 H