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Nesterboy [21]
3 years ago
12

Write a nuclear equation for the alpha decay of 23892U.23892U → 0−1e + 23893Np23892U → 10n + 23792U23892U → 42He + 23490Th23892U

→ 0−1e + 23891Pa23892U → 0+1e + 23891Pa
Chemistry
1 answer:
romanna [79]3 years ago
5 0

Answer:

23892U=23490Th +42He

Explanation:

In alpha decay, the daughter nucleus is two units less than the parent in atomic number. The mass number also decreases by 4 units. The daughter is thus found two places before the parent in the periodic table.

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Turn on Write equation. What you see is an equation that shows the original uranium atom on the left. The boxes on the right rep
My name is Ann [436]

Answer:

Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.

Explanation:

Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:

_{92} ^{238} U_{}→ _{90} ^{234} Th_{} + _{2} ^{4} He_{}

I hope this explanation is clear and explanatory.

6 0
3 years ago
How many grams of water, H20 are needed if 88<br> grams of CO2 gas are produced?
joja [24]

Answer:

Explanation:

If one mole of carbon monoxide has a mass of 28.01 g and one mole of carbon dioxide has a mass of 44.01 g , it follows that the reaction produces 44.01 g of carbon dioxide for every 28.01 g of carbon monoxide.

3 0
2 years ago
One of the products of a fermentation reaction is
Alex73 [517]
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5 0
3 years ago
N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
2 years ago
Balance NaOH + H3PO4-&gt;H2O + Na3PO4
Masteriza [31]

Answer:

3NaOH + H3PO4 ---> 3H2O + Na3PO4

Explanation:

7 0
3 years ago
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