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3 years ago

Would you expect a Boron (B) atom with atomic mass = 9.999 to be stable?

Chemistry

1 answer:

charle [14.2K]3 years ago

6 0

The balance of forces allows to find the result for the question if the isotope of boron 9.99 una is stable:

The boron isotope of mass 9.99 uma is unstable because the repulsive force increases.

The stability of atomic nuclei depends on the balance the force is electrostatic repulsion between the protons and the strong interaction of attraction.

One way to achieve this balance is to increase the separation of the protons with uncharged particles between them, the neutral ones, the strong interaction is of the same magnitude for protons and neutrons, therefore the repulsion is decreased and the strong attraction interaction is maintained. .

In the case of Boron, which has 5 protons, the stable structures have more atomistic 10 and 11 una, which is why it has 5 and 6 neutrons each. Therefore each proton has a neutrons next to it and in the other case a proton at the end has two neutrons, this causes the distance between the protons to increase, decreasing the electrostatic repulsion.

It indicates that we have a Boron nucleus of mass 9.999. The number of protons must remain fixed, therefore there are only 4 neutrons.

Consequently, some of the protons does not have a neutron next to it and can approach the other proton, therefore the electrostatic repulsion increases and the stability of the atom decreases.

In conclusion, using the balance of Forces we can find the result for the question if the isotope of boron 9.99 una is stable:

The boron isotope of mass 9.99 amu is unstable because the repulsive force increases.

Learn more about nuclear stability here: brainly.com/question/897383

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If the methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions, what volume w

Sunny_sXe [5.5K]

116.6 x $10^{ -3$ } is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C that was extracted and placed under STP conditions.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

The solubility of methane in water at 25 degrees Celcius is 1.3 x $10^{ -3}$ M. It implies 1.3 x $10^{ -3}$ moles of methane are dissolved in one litre of water.

The number of moles of methane in 4 L of water can be calculated as follows:

Moles of methane = $\frac{1.3 . 10^{ -3} }{1L}$ x $\frac{4L}{1L}$

Moles of methane = 5.2 x $10^{ -3}$

STP refers to standard temperature and pressure. Under STP conditions, the temperature of the substance is 0 degrees celcius and its pressure is 1 atm.

An ideal gas is an imaginary gas comprising of a large number of randomly moving particles and the motion between such articles is considered to be perfectly elastic. The ideal gas equation describes the relationship between pressure, volume, temperature and number of moles of a gas.

The expression for ideal gas equation is as follows:

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1 atm

V= ?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=273K

n=?

Putting value in the given equation:

$\frac{PV}{RT}=n$

5.2 x $10^{ -3} = \frac{1 \;atm\; X \;V}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 273}$

V= 116.6 x $10^{ -3}$

Hence, 116.6 x $10^{ -3$ } is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions.

Learn more about the ideal gas here:

brainly.com/question/27691721

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