Give the following number of protons, neutrons, and electrons in the following isotope:

Chemistry

1 answer:

valina [46]3 years ago

3 0

Answer:

47, 68, 47

Explanation:

Let's consider the following isotope: ¹¹⁵Ag.

If we look in the Periodic Table, the atomic number of Ag is 47, that is, silver has 47 protons.

Since ¹¹⁵Ag is neutral it must have the same number of protons that of electrons. Then, it also has 47 electrons.

The number 115 in ¹¹⁵Ag stands for the mass number, which is the sum of protons and neutrons. Then, the number of neutrons is:

p⁺ + n⁰ = 115

n⁰ = 115 - p⁺ = 115 - 47 = 68

The number of protons, neutrons, and electrons in ¹¹⁵Ag is: 47, 68, 47.

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Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2

Flauer [41]

Answer:

$\boxed{28.81}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 58.12 44.01

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g: 9.511

1. Moles of C₄H₁₀

$\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}$

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

$\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}$

3. Mass of CO₂

$\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}$