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Explanation: i took the tet
Answer:
3.79 x 10²⁴atoms
Explanation:
Given parameters:
Number moles of lead = 6.3moles
Unknown:
Number of atoms = ?
Solution:
To solve this problem,
1 mole of a substance contains 6.02 x 10²³ atoms
6.3 mole of Pb will contain 6.3 x 6.02 x 10²³ = 3.79 x 10²⁴atoms
Answer:
The question is incomplete and confusing.
- In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.
- In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.
See below the details.
Explanation:
Which compound will not form ions?
<u />
<u>1. Write the balanced molecular equation:</u>
- Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)
<u />
<u>2. Write the ionizations for the ionic aqueous compounds:</u>
<u />
- Pb(NO₃)₂(aq) → Pb⁺²(aq) + 2NO₃⁻(aq)
- 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)
- 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)
<u />
<u>3. Write the complete ionic equation:</u>
Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.
All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.
It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.
This hypothetical process would produce actinium-230.
<h3>Explanation</h3>
An alpha decay reduces the atomic number of a nucleus by two and its mass number by four.
There are two types of beta decay: beta minus β⁻ and beta plus β⁺.
The mass number of a nucleus <em>stays the same</em> in either process. In β⁻ decay, the atomic number <em>increases </em>by one. An electron e⁻ is produced. In β⁺ decay, the atomic number <em>decreases </em>by one. A positron e⁺ is produced. Positrons are antiparticles of electrons.
β⁻ are more common than β⁺ in decays involving uranium. Assuming that the "beta decay" here refers to β⁻ decay.
Gamma decays do not influence the atomic or mass number of a nucleus.
Uranium has an atomic number of 92. 238 is the mass number of this particular isotope. The hypothetical product would have an atomic number of 92 - 2 ⨯ 2 + 1 = 89. Actinium has atomic number 89. As a result, the product is an isotope of actinium. The mass number of this hypothetical isotope would be 238 - 2 ⨯ 4 = 230. Therefore, actinium-230 is produced.
The overall nuclear reaction would involve five different particles. On the reactant side, there is
On the product side, there are
- one actinium-230 atom,
- two alpha particles (a.k.a. helium-4 nuclei),
- one electron, and
- one gamma particle (a.k.a. photon).

Consider: what would be the products if the nucleus undergoes a β⁺ decay instead?