1pt. An atom of carbon-14 contains six protons, eight neutrons, and six electrons. Which will an atorn of carbon-12 contain? O A. seven protons, five neutrons, and six electrons
OB. six protons, six neutrons, and six electrons
OC. five protons, seven neutrons, and five electrons
OD. 12 protons, no neutrons, and 12 electrons

B. six protons, six neutrons, and six electrons

Explanation:

Carbon-14 and Carbon-12 are the isotopes of carbon.

Isotope:

An atom of an element that contain same atomic number but different atomic mass.

The difference in atomic mass is due to the different number of neutrons present, but the number protons and electrons are remain same.

In C-14 there are six protons and six electrons while number of neutrons are,

Mass number - number of protons  = number of neutrons

14- 6 =  8

In C-12 there are six protons and six electrons while number of neutrons are,

Mass number - number of protons  = number of neutrons

12- 6 =  6

Thus in C-12, there are six protons, six neutrons and six electrons present.

OB. six protons, six neutrons, and six electrons

Explanation:

Atoms of carbon - 14:                      Carbon -12

              Protons  = 6                                6

              Neutrons  = 8                             6

              Electrons  = 6                               6

An atom of carbon-12 will have an equal number of protons and electrons but different number of neutrons. The two give species are isotopes.

Therefore, the two isotopes will differ in the number of neutrons they have.

your answer would be “dry”

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.