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4 years ago

What does it mean to weaponize or enrich uranium

2 answers:

4 0

Enriched uranium is a type of uranium in which the percent composition of uranium-235 has been increased through the process of isotope separation. Natural uranium is 99.284% 238U isotope, with 235U only constituting about 0.711% of its mass. ... The term oralloy is still occasionally used to refer to enriched uranium. Hope this helps

fredd [130]4 years ago

3 0

Enriched uranium is a kind of uranium where the percent composition of uranium-235 has been increased through the process of isotope separation, Natural uranium is 99.284% 238U isotope, with 235U only constituting about 0.711% of its mass

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Which observations is the best evidence that unknown liquid X is a composed of only pure substance

Angelina_Jolie [31]

Answer:

A substance that is composed only of atoms having the same atomic number is ... 36 grams of an unknown liquid at its boiling point,.

4 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

4 years ago

B. It shifts the equilibrium toward the right, favoring product.

PilotLPTM [1.2K]

Answer:

11) the difference in heat energies between products and reactants

12) enthalpy change

Explanation:

The heat of reaction is defined as that energy released or absorbed as chemical substances participate in a chemical reaction. It is a term used to denote the change in energy as reactants change into products.

Another name of heat of reaction is enthalpy of reaction. It is a state function since it depends on the initial and final states of the system.

8 0

3 years ago

N the ionic compound aluminum selenide, each atom of aluminum (gains or loses)

Volgvan

Answer:

loses, gains

Explanation:

In the ionic compound aluminum selenide, each atom of aluminum will lose electrons while each atom of selenium will gain the electrons.

An ionic compound is an interatomic bond formed between a metal and non-metal. The metal is less electronegative compared to the non-metal. In this case, the metal will lose electrons to become positively charged whereas the non-metal, selenium will gain the electron to become negatively charged.

The electrostatic attraction between these oppositely charged ions leads to the formation of the ionic bond.

6 0

3 years ago

Which element will have the lower ionization energy? Be. Ba

sp2606 [1]

Boron would have the lower first ionization energy.
Be: 1s2 2s2
B: 1s2 2s2 2p1 -> electron from the p orbital is easier to eject than than electron from the s orbital.

6 0

4 years ago

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