When an isotope emits an alpha particle, its atomic number

Chemistry

1 answer:

Yuri [45]3 years ago

5 0

Answer:

The atomic number of the isotope is reduced by 2 and atomic mass is reduced by 4.

Explanation:

Alpha particle is doubly ionized helium atom. The atomic mass of helium atom is 4 u and atomic number of helium atom is 2. The emission of alpha particle means doubly ionized helium atom will be released from the isotope.

Hence, the emission of the alpha particle will decrease the atomic mass by 4 and atomic number by 2.

Hence, option(a) is correct

You might be interested in

Given 2.91 moles of a gas in a 500 milliliter-container, if the temperature is found to be 31 degrees Celsius, what is the press

kari74 [83]

Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K

P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm

3 0

3 years ago

Read 2 more answers

Amplitude measures the light's

charle [14.2K]

Answer:

d.) intensity

Explanation:

8 0

2 years ago

Read 2 more answers

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser: