According to the kinetic-molecular theory, what happens to a liquid when it is transferred from one container to another?
a.) the volume and the shape stay the same.
b.) the volume increases to fill the new container, but the shape stays the same.
c.) the volume stays the same, but the shape changes to fit the new container.
d.) the volume and the shape change to fill the new container.

C.) The volume stays the same, but the shape changes to fit the new container.

Explanation: There are three states of matter namely solid, liquid and gas.

Solids have fixed shape and fixed volume as the molecules are strongly bonded to each other. In this state, the molecules are arranged in regular and repeating pattern.The molecules are closely packed that means they are fixed and vibrate in place but they can not move from one place to another.

Liquids have fixed volume but no fixed shape as the molecules are weakly bonded to each other. Liquids acquire the shape of the container in which it is put. In this state, the molecules are present in random and irregular pattern. The molecules are closely packed but they can move from one place to another.

Gases have neither fixed shaped nor fixed volume as the molecules are very weakly bonded to each other. In this state, the molecules are present in irregular pattern. The molecules are not closely packed and they can move freely from one place to another and spread out.

hi,

the answer is d, it has indefinite shape but definite volume.

hope this !

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