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3 years ago

Boron occurs naturally as two isotopes. What is the difference between these isotopes?

2 answers:

Citrus2011 [14]3 years ago

6 0

"They have different numbers of neutrons and different mass numbers. Remember mass number is protons plus neutrons. The number of protons for an element will never change, however the number of neutrons can." I took the test, its D

Oksanka [162]3 years ago

4 0

Answer: The two isotopes of boron differs in mass number rather than atomic number.

Explanation:

Isotopes are defined as the chemical species that have same atomic number but differ in their mass number.

Boron is the 5th element of the periodic table and have 2 naturally occurring isotopes. The isotopes are : B-10 and B-11

For $_5^{10}\textrm{B}$ isotope:

Percentage abundance of this isotope = 20 %

Atomic number = 5

Mass number = 10

For $_5^{11}\textrm{B}$ isotope:

Percentage abundance of this isotope = 80 %

Atomic number = 5

Mass number = 11

Hence, the two isotopes of boron differs in mass number rather than atomic number.

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What is radioactive decay?

swat32

Answer:

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive. Three of the most common types of decay are alpha decay, beta decay, and gamma decay, all of which involve emitting one or more particles or photons.

Explanation:

7 0

2 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

The single bond is

Alex17521 [72]

4 is correct for this one

3 0

3 years ago

Read 2 more answers

c. The reaction Br2 (l) --> Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a

egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0

3 years ago

What is the energy of a photon with a frequency of 9.50 x 1013 Hz?

Taya2010 [7]

Answer:

$\huge{6.295 \times {10}^{ - 20} \: J}$

Explanation:

The energy of the photon can be found by using the formula

E = hf

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question we have

E = 6.626 × 10-³⁴ × 9.5 × 10¹³

We have the final answer as

$6.295 \times {10}^{ - 20} \: J$

Hope this helps you

3 0

2 years ago