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2 years ago

What are the symbols, atomic mass and abundance in nature of the isotope boron-11

Chemistry

1 answer:

Aneli [31]2 years ago

8 0

The chemical symbol of the isotope boron-11 is ¹¹B.
The atomic mass of the isotope boron-11 is equal to 11.009306.
The abundance in nature of the isotope boron-11 is equal to 80.1%.

<h3>What is an isotope?</h3>

An isotope can be defined as the atom of a chemical element that has the same number of protons but different number of neutrons. This ultimately implies that, the isotopes of an element have the same atomic number (number of protons) but different atomic mass (number of neutrons).

In Chemistry, there are two main isotopes of boron and these include the following:

Boron-10

Boron-11

Boron-11 is the most stable isotope of boron and it is characterized by the following:

The chemical symbol of the isotope boron-11 is ¹¹B.
The atomic mass of the isotope boron-11 is equal to 11.009306.
The abundance in nature of the isotope boron-11 is equal to 80.1%.

Read more on Boron-11 here: brainly.com/question/6283234

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Which statement describes the energy changes that occur as bonds are broken and formed during a chemical reaction? 1.Energy is a

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3 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

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Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

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3 years ago

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Answer:

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A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

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4 0

3 years ago