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3 years ago

Sort these isotopes by whether they are most likely to undergo fusion or fission.

Chemistry

1 answer:

Maslowich3 years ago

4 0

Answer:

Uranium-233- fission

Plutonium-239- fission

Plutonium-241- fission

Hydrogen-3 fusion

Hydrogen-1 fusion

Helium-3 fusion

Explanation:

In nuclear fission, heavy nuclear disintegrate into smaller nuclei when bombarded with particles such as neutrons. Fission reaction is common among nuclei having a high atomic number such as plutonium and uranium.

Fusion occurs between two light nuclei such as hydrogen or helium. It involves the combination of two lighter elements to give a heavier element with the release of tremendous amount of energy.

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Calculate the pressure exerted by Ar for a molar volume of 0.590 L⋅mol−1 at 295 K using the van der Waals equation of state. The

Volgvan

Answer:

the pressure P= 40.03 bar

Explanation:

The Van der Waals equation states that

P= R*T/(V-b) - a/V²

where

T= absolute temperature = 295 K

V= molar volume = 0.590 L/mol

R= ideal gas constant = 0.082 atm*L/mol*K

a= van der Waals parameter = 1.355 bar⋅dm⁶/mol² * (0.987 atm/bar) * (1 L²/dm⁶) = 1.337 atm ⋅L²/mol²

b= van der Waals parameter = 0.032 dm³/mol *(1 L/dm³) = 0.032 L/mol

then replacing values

P= R*T/(V-b) - a/V² = 0.082 atm*L/mol*K*295 K/( 0.590 L/mol-0.032 L/mol) - 1.337 atm ⋅L²/mol²/(0.590 L/mol)² = 39.51 atm

P= 39.51 atm / (0.987 atm/bar) = 40.03 bar

8 0

3 years ago

A 25.00 ml sample of an aqueous solution of ba(oh)2 requires 18.45 ml of 0.01500 m hcl (aq) for its neutralization. what is the

klio [65]

5.54 x 10⁻³ M

<h3>Further explanation</h3>

Given:

Neutralization reaction between:

25.00 ml of Ba(OH)₂
18.45 ml of 0.015 M HCl

Question:

What is the molarity of the Ba(OH)₂ solution?

The Process:

Let us say the molarity of the Ba (OH)₂ solution as x M.

Step-1: prepare moles for each reagent

$\boxed{ \ M = \frac{n}{V} \ }$ $\rightarrow \boxed{ \ n = MV \ }$

$\boxed{ \ Ba(OH)_2 \rightarrow n = (x \ \frac{mol}{L})(25 \ ml) = 25x \ mmol \ }$

$\boxed{ \ HCl \rightarrow n = (0.015 \ \frac{mol}{L})(18.45 \ ml) = 0.277 \ mmol \ }$

Step-2: neutralization

We use the ICE table to see how neutralization occurs between acid and base.

Balanced reaction:

$\boxed{ \ Ba(OH)_2_{(aq)} + 2HCl_{(aq)} \rightarrow BaCl_2_{(aq)} + 2H_2O_{(l)} \ }$

Initial: 25x 0.277 - -

Change: - ¹/₂ · (0.277) -0.277 +¹/₂ · (0.277) +0.277

Equlibrium: - - +¹/₂ · (0.277) +0.277

Neutralization causes no excess of hydrogen or hydroxide ions in solution. In the end, the number of acid and base reactions is balanced. In other words, the two reagents have run out with nothing left.
HCl acts as a limiting reagent.

Step-3: calculate the molarity of the Ba(OH)₂ solution.

We consider Ba (OH) from the initial, change, and equilibrium stages.

$\boxed{ \ 25x - \frac{1}{2}(0.277) = 0 \ }$

$\boxed{ \ 25x = \frac{1}{2}(0.277) \ }$

$\boxed{ \ 50x = 0.277 \ }$

$\boxed{ \ x = \frac{0.277}{50} \ }$

$\boxed{ \ x = 5.54 \times 10^{-3} \ }$

Thus, the molarity of the Ba(OH)₂ solution is 5.54 x 10⁻³ M.

_ _ _ _ _ _ _ _ _ _

Alternative Steps

Valence of base = the number of OH⁻ ions
Valence of acid = the number of H⁺ ions

Neutralization: $\boxed{ \ V_b \cdot M_b \cdot valence \ of \ base = V_a \cdot M_a \cdot valence\ of \ acid \ }$

$\boxed{ \ (25.00 \ ml) \cdot x \cdot 2 = (18.45 \ ml) \cdot (0.015 \ M) \cdot 1 \ }$

$\boxed{ \ x = \frac{(18.45 \ ml) \cdot (0.015 \ M)}{(25.00 \ ml) \cdot 2} \ }$

Thus the same results were obtained. The molarity of Ba (OH) ₂ solution is 5.54 x 10⁻³ M.

<h3>Learn more</h3>

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