Which set of atoms are isotopes? A. 16^ 8X and 16^ 6X
B. 16^ 8X and 17^ 8X
C. 12^ 6X and 14^ 7X
D. 6^ 26X and 26^ 6X

Option B.

Explanation:

Isotopy is a phenomenon where by two or more atoms have the same atomic number (proton) but different mass number due to the difference in neutron present in the atoms. The atoms that exhibit Isotopy are isotopes.

With the above information, we can obtain the right answer to the question given above as follow:

For option A:

Isotope >> atomic no >> Mass no

16 8X >>> 8 >>>>>>>>> 16

16 6X >>> 6 >>>>>>>>> 16

Since they have the different atomic number, they are not isotope.

For option B:

Isotope >> atomic no >> Mass no

16 8X >>> 8 >>>>>>>>> 16

17 8X >>> 8 >>>>>>>>> 17

Since they have the same atomic number but different mass number, they are isotope.

For option C:

Isotope >> atomic no >> Mass no

12 6X >>> 6 >>>>>>>>> 12

14 7X >>> 7 >>>>>>>>> 14

Since they have different atomic number, they are not isotope.

For option D:

Isotope >> atomic no >> Mass no

6 26X >> 26 >>>>>>>>> 6

26 6X >>> 6 >>>>>>>>> 26

Since they have different atomic number, they are not isotope.

From the above illustration, only option B contains isotopes because the atoms have the same atomic number but different mass number

answer: i would say b.

explanation:

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