Answer:
The molarity of the lead(II) ion in the original solution is 0.03M
Explanation:
<u>Step 1:</u> The balanced reaction
Pb(NO3)2(aq) + 2 NaI aq) → PbI2(s) + 2 NaNO3 (aq)
Pb2+ + 2I- →PbI2
This means that for 1 mole Pb2+ consumed, there is needed 2 moles of I- to produce 1 mole of PbI2
<u>Step 2:</u> Calculate moles of PbI2
Moles of PbI2 = 0.862g / 461.01 g/mole
moles of PbI2 = 0.00187 moles
<u>Step 3:</u> calculate moles of Pb2+
For 1 mole of PbI2 produced we need 1 mole of Pb2+
This means for 0.00187 moles of PbI2 we need 0.00187 moles of Pb2+
<u>Step 4:</u> Calculate the molarity of the Pb2+ ion
Molarity of Pb2+ = moles of Pb2+ / volume =
Molarity of Pb2+ = 0.00187 moles / 62.3*10^-3
Molarity of Pb2+ ion = 0.03M
The molarity of the lead(II) ion in the original solution is 0.03M
Answer:
option 3 = 3.41 m
Explanation:
Given data:
Frequency of signals = 88.1 MHz or 88.1 × 10⁶ Hz
Wavelength = ?
Solution:
Formula:
λ = c / f
λ = Wavelength
c = speed of light
f = frequency
λ = c / f
λ = 3 × 10⁸ m.s⁻¹ / 88.1 × 10⁶ s⁻¹
λ = 0.0341 × 10² m or 3.41 m
Answer:
A. the temperature at which the motion of particles theoretically ceases.
Explanation:
Absolute zero is the same as 0 K (or zero Kelvin). At this point, the temperature cannot get any lower, and it is incredible difficult to get something to this low low temperature. The particles theoretically stop moving at 0 Kelvin aka absolute zero.
Answer:
a. syn
b. E1
c. E2
d. E2
Explanation:
In organic chemistry, the rule of Saytzeff, Saytzev or Zaitsev states that in an elimination reaction (β-elimination) in which more than one alkene can be formed, the most thermodynamically stable will be the majority.
In general, the most substituted alkene is the most stable, due to the electronic sharing properties of the alkyl groups with the double bond C = C (hyper-conjugation). Also, in some cases, another stabilizing effect may be incurred when establishing the regioselectivity such as the conjugation of the double bond with other groups.
This rule is valid except in bimolecular elimination reactions (E2) in which there is a significant steric hindrance, branched substrate and / or bulky base, and without the possibility of conjugation, then applying Hofmann's rule.
The elimination reaction E1 has a potential energy profile similar to that of an SN1 reaction. The formation step of carbocation is very endothermic, with a transition state which is what determines the speed of the reaction. The second step is a rapid and exothermic deprotonation. Base does not participate in the step that determines the speed of the process, so it only depends on the concentration of alkyl halide.
The bimolecular elimination E2 takes place without intermediates and consists of a single ET, in which the base abstracts the proton, the leaving group leaves and the two carbons involved are rehybridized from sp3 to sp2.