Don’t know need the anwser

The distance is slightly less than 3.5 m

Explanation: assuming wall and target are the same thing, and the bullet has constant velocity, the bullet will travel 7 m in half a second, so half that distance is 3.5 m.

In reality, the bullet is decelerating (at an unknown rate) so the distance is slightly less than 3.5 m.

There is also a vertical velocity component, which means it hits the target/wall at an angle. The trajectory is such that it hits the wall above the shooter because the ricochet hits at ~the level at which it left the firearm.

If the wall was absent, the bullet would have described a parabola which brough it back to the initial level after 7m. This could be calculated, but it means that the actual distance between the shooter and the wall is slightly less than 3.5 m

In addition, the collision with the wall is not 100% elastic, so the velocity aftercthe ricochetvis further reduced.

A calculation would be complex because these confounding factors are not completely independent of each other,  but all reduce the average velocity and therefore the distance.

Therefore it is only possible to say that the distance was somewhat less than 3.5 m

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