The entropy S(U, V) is a function of U and V in equilibrium. Using the chain rule, dS(U, V) = (dS(U, V)/dU)V dU + (dS(U, V)/dV)U dV
This is always true; it's a mathematical identity.
The "Fundamental Relation of Thermodynamics" says that in equilibrium:
TdS = dU + pdV
where T is temperature, S is entropy, U is internal energy, p is pressure, and V is volume. This describes how, in equilibrium, S depends on changes in U and V.
Here are other derivatives to evaluate, always assuming equilibrium so you can use the Fundamental Relation
1) What's (dS(U, V)/dV)U?
2) What's (dS(U, V)/dU)V?
3) What's (dS(S, V)/dV)S?

a graduated cylinder.

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

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.