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2 years ago

Calculate the mass defect of the boron nucleus 11 5b. The mass of neutral 11 5b is equal to 11.009305 atomic mass units.

Chemistry

2 answers:

yan [13]2 years ago

8 0

The mass defect of the Boron : <u>Δm = 0.081815 amu</u>

<h3>Further explanation</h3>

Atoms are composed of 3 types of basic particles namely protons, electrons, and neutrons

The mass of this basic particle is expressed in atomic mass units (amu)

The mass of atoms is always slightly less than the sum of the masses of the individual atomic forming particles namely neutrons, protons, and electrons. The difference between the mass of the atom and the sum of the masses of atomic subparticles is called the mass defect (Δm).

Can be formulated:

Δm = [Z (mp + me) + (A-Z) mn] - m atom

where:

Δm = mass defect (amu)

mp = mass of a proton (1,007277 amu)

mn = mass of a neutron (1.008665 amu)

me = mass of an electron (0.000548597 amu)

matom = mass of nuclide (amu)

Z = atomic number

A = mass number

Atomic number (Z) = number of protons = number of electrons

Mass Number (A) is the sum of protons and neutrons

Atomic Number (Z) = Mass Number (A) - Number of Neutrons

For Boron elements (₁₁⁵B) which have atomic number 5 and mass number 11, then:

number of protons = 5

number of electrons = 5

number of neutrons = mass number - atomic number = 11 - 5 = 6

then:

Δm = [5 (1.007277 + 0.000548597) + (6) 1.008665] - 11.009305

Δm = [5 (1.007826) + (6) 1.008665] - 11.009305

Δm = (5.03913 + 6.05199) - 11.009305

Δm = 11.09112 - 11.009305

Δm = 0.081815 amu

<h3>Learn more </h3>

element symbol

brainly.com/question/2572495

the chemical symbols

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The subatomic particle that has the least mass

brainly.com/question/2224691

Keywords: mass number, atomic mass, amu, mass defect, atomic number

Vlad1618 [11]2 years ago

7 0

Answer:

0.07906687 amu

Explanation:

For Boron ₅B¹¹, the number of protons is 5 and the mass is 11. The mass is the number of protons plus the number of neutrons, so:

neutrons = 11 - 5 = 6

The mass of an atom is concentrated in the nucleus, so it is the mass of the protons + the mass of the neutrons. The mass of 1 proton is 1.00727647 amu/proton, and the mass of 1 neutron: 1.00866492 amu/neutron, so for the element given the theoretical mass (mt) is:

mt = 5* 1.00727647 amu/proton + 6*1.00866492 amu/neutron

mt = 11.08837187 amu

The mass defect (md) is the theorical mass less the real mass:

md = 11.08837187 - 11.009305

md = 0.07906687 amu

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1 year ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

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2 years ago

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached

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Answer:

(a) Cu²⁺ +2e⁻ ⇌ Cu

Explanation:

(a) Cu half-reaction

Cu²⁺ + 2e⁻ ⇌ Cu

(c) Cell voltage

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Cu²⁺ + 2e⁻ ⇌ Cu; 0.34

Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241

The equation for the cell reaction is

E°/V

Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34

Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10

The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation

(ii) Calculations:

T = 25 + 273.15 = 298.15 K

$Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}$

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2 years ago

The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil

jasenka [17]

Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Given data

Average distance between nitrogen and oxygen atoms: 115 pm

Step 2: Convert the distance to meters (SI base unit)

We will use the conversion factor 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Convert the distance to millimeters

We will use the conversion factor 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

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2 years ago