Imagine that you heat a pot of water to a boil (212°F), then you turn off the stove and let the water sit there. The room is at room temperature (72°F) What temperature do you predict the water cool to after sitting for 8 hours? (

72 °F

Explanation:

Hello,

In this case, since matter tend the be stable at conditions they do not change, if we have water at room temperature (72 °F) and we heat it until boiling but then we turn off the stove, we molecularly see that due to the heat, the molecules start moving faster and crashing to each other; however when that fed energy starts being removed as the stove gets turned off, the molecules lose energy causing them to start moving slower so they lose energy until a point the remain at the initial status, that is 72 °F because thermal equilibrium is reached. Therefore it is likely for the water to be at 72 °F as well once 8 hours have passed, because a thermal equilibrium is reached with the environmetnal temperature.

Best regards.

c is the answer because when something is left after something evaporates that shows the chemical change.

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