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ki77a [65]
2 years ago
8

What isotope is formed when u-238 emits an alpha particle?.

Chemistry
1 answer:
yarga [219]2 years ago
4 0

A nucleus of uranium 238 decays by alpha emission to form a daughter nucleus, thorium 234.

<h3>What are isotopes?</h3>

Isotopes are atoms with the same number of protons but differ in numbers of neutrons.

Uranium-238 produces thorium-234 by alpha decay.

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, uranium-238 decays through α-particle emission to form thorium-234 according to the equation:

^{238}_{92}U\;\rightarrow\;^{234}_{90}Th\;+\;^4_2He

Learn more about the isotopes here:

brainly.com/question/11680817

#SPJ1

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Which one of the following is no equal to 100 meters
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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
37.5g/mL converted to kg/L
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4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.
slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ\frac{1mol}{102g} = <em>0,0980 moles</em>

And 10,0g of HCl are:

10,0 gₓ\frac{1mol}{36,5g} = <em>0,274 moles</em>

<em />

For a total reaction of 0,274 moles of HCl you need:

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Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl ×\frac{2 moles AlCl_{3}}{6 moles HCl} × 133\frac{g}{mol} = <em>12,1 g of AlCl₃</em>

<em />

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles

And its mass is:

0,0523 molesₓ\frac{102g}{1mol} = <em>5,33 g of Al₂O₃ </em>

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I hope it helps!

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