A)a concentration c(mol/l) varies with time (min) according to the equation c=3.00exp(−2.00t) a) what are the implicit units of 3.00 and 2.00? b) suppose the concentration is measured at t = 0 and t = 1 min. use two-point linear interpolation or extrapolation to estimate c(t=0.6min) and t(c=0.10mol/l) from the measured values, and compare these results with the true values of these quantities. c) sketch a curve of c versus t, and show graphically the points you determined in part (b).

a. 3.00 must be in mol/L and 2.00 in 1/min.

b. Using lineal equation:

t = 0.6 min then C = 1.45 mol/L

C = 0.10 mol/L then t = 1.12 min

Using exponential equation:

t = 0.6 min then C = 0.90 mol/L

C = 0.10 mol/L then t = 1.70 min

Explanation:

a) C is in mol/L so, as we know that the exponential does not have units, 3.00 must be in mol/L and 2.00 in 1/min.

b) If we want to make a line between 0 an 1 we need to find the linear equation for this.

We can use the slope intercept-equation:

y = ax + b (1)

Let's find a and b.

a is the slope of here, so we can use the next equation to find it:

We have two points, (0,C₁) and (1,C₂)

So a will be:

We can use the equation (1) to find b, for instance, let's choose:

Finally, our linear equation will be: or  

Using lineal equation:

t = 0.6 min then C = 1.45 mol/L

C = 0.10 mol/L then t = 1.12 min

Using exponential equation:

t = 0.6 min then C = 0.90 mol/L

C = 0.10 mol/L then t = 1.70 min

I hope it helps you!

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever