A total of 620 C of charge passes through a flashlight in 0.510 h. What is the average current in milliamperes?

Explanation:

The given data is as follows.

               Q = 620 C,         t = 0.510 hour

First, we will convert hours into seconds. Since, there are 3600 seconds present in 1 hour. Hence, number of seconds present in 0.510 hour are as follows.

                     = 1836 sec

Therefore, we will calculate the average current as follows.

                      I =

                        =

                        = 0.337 C/sec

It is known that,

                      1 C/sec = 1000 mA

So,            

                    = 337 mA

Thus, we can conclude that the average current in milliamperes is 337 mA.

the answer is c i believe

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