14C is an isotope of carbon that possesses: a) 6 protons, 8 neutrons, and 2 electrons. b) 6 protons, 6 neutrons, and 2 electrons. c) 6 protons, 8 neutrons, and 6 electrons. d) 8 protons, 6 neutrons, and 2 electrons. e) 6 protons, 2 neutrons, and 6 electrons.

Option C. 6 protons, 8 neutrons and 6 electrons.

Explanation:

14C has the following:

Mass number = 14

Atomic number = 6

Next we shall determine the number of protons in 14C. This can be obtained as follow:

Atomic number is simply the number protons in an atom. Thus,

Proton = Atomic number

Atomic number = 6

Proton = Atomic number = 6

Proton = 6

Next, we shall determine the number of neutrons in 14C. This can be obtained as follow:

Mass number = proton + neutron

Mass number = 14

Proton = 6

Neutron =.?

Mass number = proton + neutron

14 = 6 + neutron

Collect like terms

Neutron = 14 - 6

Neutron = 8

Finally, we shall determine the number of electrons in 14C.

This can be obtained as follow:

Since the isotope i.e 14C has no charge, it means that the atom is in it's ground state. Hence, the number of protons and the number of electrons are the same i.e

Electron = proton

Proton = 6

Electron = proton = 6

Electron = 6

From the above calculations, 14C has

6 protons, 8 neutrons and 6 electrons.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

it takes on an electronical energy