Breeder reactors are used to convert the nonfissionable nuclide U-238 to a fissionable product. Neutron capture by the U-238 is
1 answer:
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Answer : The correct option is, (D) 89.39 KJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH
Thus, the number of neutralized moles = 0.3562 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 274 g
= final temperature of water = 325.8 K
= initial temperature of metal = 298 K
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -31.84 KJ
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -31.84 KJ
n = number of moles used in neutralization = 0.3562 mole
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 89.39 KJ/mole
<h3>Answer:</h3>
2 L Ne
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Step 1: Define</u>
0.07 mol Ne (g)
<u>Step 2: Identify Conversions</u>
STP - 22.4 L per mole
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
1.568 L Ne ≈ 2 L Ne