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4 years ago

The mass of an atom is ________________________.

Chemistry

2 answers:

tigry1 [53]4 years ago

5 0

A. distributed uniformly throughout the atomic sphere

I think it might be right

Thepotemich [5.8K]4 years ago

4 0

Chur answerwould be B concentrated in the nucleus

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Study the graphs

oee [108]

Answer:

Graph A

Explanation:

4 0

3 years ago

What is atom.

katovenus [111]

Answer:

atom is the smallest indivisible particle of an element

Explanation:

The symbol of 20 element are H He li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca

element is the simplest pure form of a substance .

3 0

2 years ago

The first excited electronic energy level of the helium atom is 3.13 x 10-18 J above the ground level. Estimate the temperature

salantis [7]

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸ J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

$\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} = e ^{^{-\dfrac{(\Delta E)}{KT}}}$

From the above equation:

Δ E = energy difference = 3.13 × 10⁻¹⁸ J

k = Boltzmann constant = 1.38 × 10⁻²³ J/K

$\dfrac{N_2}{N_1} = 0.5$

Thus:

$0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = -226811.5942 \times \dfrac{1}{T}$

$T = \dfrac{-226811.5942}{-3.00 }$

T = 75603.86473 K

4 0

3 years ago

90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$

ΔᵣH° is negative, so the reaction is exothermic.

5 0

3 years ago

What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?

Sunny_sXe [5.5K]

Answer:

2,67 L

Explanation:

8 0

2 years ago