How did scientists refute the early atomic theory claim that atoms were indivisible?
select all that apply.
a. scientists bombarded atoms of large mass with neutrons, observed reaction products with slightly smaller mass, and described this result as fusion.
b. scientists collided atoms of small mass together, observed reaction products of slightly larger mass, and described this result as fusion.
c. scientists bombarded atoms of large mass with neutrons, observed reaction products with much smaller mass, and described this result as fission.
d. scientists collided atoms of small mass together, observed reaction products with twice the mass of original atoms, and described this result as fusion.

D. Scientists collided atoms of small mass together, observed reaction products with twice the mass of original atoms, and described this result as fusion.

Explanation:

The early atomic theory was purposed by Dalton and said that the atoms are the smallest particle, so they were indivisible. Further theories refuted that theory, such as Rutherford's experiment, in which particles alpha was bombarded in a gold foil, and was noticed that some of then have deviated, most passed through, and a few deviated completed.

Rutherford purposed that the atom is an empty space, with a small nucleus in the middle, and electrons orbiting it. But, most recently, by bombarding atoms of small mass together (such as hydrogen), scientists observed that other atoms with the double mass were formed, which means that it is impossible to the atom to be indivisible because had happened a fusion.

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

answer; natural resources;