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natta225 [31]
2 years ago
9

Four atoms have the numbers of protons and neutrons given below.

Chemistry
1 answer:
Makovka662 [10]2 years ago
3 0

Answer:

W and X

Explanation:

isotopes : are masses of atoms with the same number of proton and electron but different neutron .

W and X have the same number of protons and different number of neutrons.

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Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t
Dvinal [7]

Answer :  The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

6.941=[(6.01512\times x)+(7.01600\times (100-x))]

By solving the term 'x', we get:

x=694.048

Percent abundance of Li isotope-1 = \frac{694.048}{100}=6.94\%

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0
2 years ago
How is the mass of carbon related to the mole?
Anuta_ua [19.1K]

Answer:

Simply put, you can go from moles to grams and vice versa by using the mass of 1 mole of that substance, i.e its molar mass. For example, the molar mass of carbon is 12.011 g/mol. This means that 1 mole of carbon, or 6.022⋅1023 atoms of carbon, weigh 12.011 g.

Explanation:

4 0
3 years ago
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. w
o-na [289]

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

4 0
3 years ago
How many bromine atoms are present in 37.5 g of CH2Br2?
ella [17]
<span>37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2

4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
5 0
2 years ago
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

= De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of beta particle = 9.1094\times 10^{-31} kg

E_k = kinetic energy of the particle = 4.71\times 10^{-15}J

Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

3 0
3 years ago
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