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alexandr402 [8]
3 years ago
14

Which of the following describes the middle layers of the earth

Chemistry
1 answer:
Feliz [49]3 years ago
8 0
What are the choices? I can still help answer but there aren’t any following answers lol
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Can you please help me on my Chemistry homework.
Yanka [14]

Answer: Hold on, our servers are swamped. Wait for your answer to fully load.

Explanation: i took the tet

3 0
2 years ago
How many atoms are in 6.3 moles of lead
densk [106]

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

4 0
4 years ago
Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Which ion would NOT be present in
Allisa [31]

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.

3 0
3 years ago
HELP PLEASE.
Dafna11 [192]

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

  • one uranium-238 atom.

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).

\;_{\phantom{2}92}^{238} \text{U} \to \;_{\phantom{2}89}^{230} \text{Ac} + \;_{2}^{4} \text{He} + \;_{2}^{4} \text{He} + \text{e}^{-} + \gamma

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

8 0
3 years ago
All molecules have the same kinetic energy and hence the same speed.<br><br> O True<br><br> O False
topjm [15]
False...their speed differs
4 0
3 years ago
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