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-Dominant- [34]
3 years ago
8

Is the following nuclear equation balanced? 237 Np - He+233 AM

Chemistry
1 answer:
Flura [38]3 years ago
6 0

Answer:

Yes

Explanation:

Np²³⁷   →  He⁴ + Am²³³

The given nuclear equation is balanced. Np²³⁷ undergoes alpha decay and produce alpha particle and Am²³³.

Properties of alpha radiation:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number 4 less than and atomic number 2 less than the starting atom.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

₉₂U²³⁸   →   ₉₀Th²³⁴  + ₂He⁴  + energy

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When the following reaction reached equilibrium at 325 K , the equilibrium constant was found to be 172. When a sample was taken
miss Akunina [59]

Answer: 6.88M

Explanation:

K= [N204]/[NO2]^2

[NO2]= 0.2M

K=172

[N204]=K * [N02]^2

172 *0.2^2

=6.88M

6 0
3 years ago
What types of objects are using electricity in your home today?
EleoNora [17]

Lamp, tv, refrigerator, Lights, phone, microwave,  surround sound, cameras, motion sensors, chandelier, pool, hot tub, heated floors, the smart home stuff,  aquarium stuff, and whatever the staff uses.

3 0
3 years ago
Write a net ionic equation to show that hydroiodic acid, hi, behaves as an acid in water.
OlgaM077 [116]

Answer:

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Explanation:

The HI donates a proton to the water, converting it to a hydronium ion

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Thus, the HI is behaving like a Brønsted acid.

3 0
3 years ago
Read 2 more answers
Select the two elements isolated by the Curies.
Mkey [24]
Radium and polonium is the answer to this question. I hope I helped out!
5 0
3 years ago
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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

4 0
3 years ago
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