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statuscvo
1 month ago
13

A tritium nucleus is formed by combining two neutrons and a proton. the mass of this nucleus is 9.106 × 10–3 universal mass unit

less than the combined mass of the particles from which it is formed. approximately how much energy is released when this nucleus is formed.
Chemistry
2 answers:
lions [2.9K]1 month ago
7 0
E = mc²

where E = energy produced

m = mass of the nucleus

C = speed of light

m = 9.106 x 10⁻³ x 1.67 x 10⁻²⁷ kg

C = 3 x 10⁸ m/s, thus C² = 9 x 10¹⁶

E = 1.37 x 10⁻¹² J

VMariaS [2.9K]1 month ago
6 0

Answer: The energy released is, 136.043\times 10^{-14}J

Explanation:

According to Einstein's formula, energy equals mass multiplied by the square of the speed of light.

The formula is:

E=m\times c^2

where,

E = energy

c = speed of light = 3\times 10^8m/s

m = mass of nucleus = 9.106\times 10^{-3}amu

Transforming the mass of the nucleus from 'amu' to 'Kg':

Thus, 1amu=1.66\times 10^{-27}Kg

So, 9.106\times 10^{-3}amu=(9.106\times 10^{-3})\times (1.66\times 10^{-27})Kg

We can now insert the provided values into the formula, leading us to the amount of energy produced.

E=[(9.106\times 10^{-3})\times (1.66\times 10^{-27})Kg]\times (3\times 10^8m/s)^2

E=136.043\times 10^{-14}J

Consequently, the energy released is, 136.043\times 10^{-14}J

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Hence, Venus and Mars are NEIGHBORS to EARTH.

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66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu
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Answer:

\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

                 2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V:            66.667

c/mol·L⁻¹:   4.000       3.000

(b) Moles of H₂SO₄

\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}

(d) Volume of Al(OH)₃

\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

 q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

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Mass of aluminium hydroxide solution =  50.000    

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(b) ΔT

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(c) ΔH

\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\

\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}

This result appears nonsensical, but it is derived from your given figures.

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