Why do we learn about Laws and Motion

1 answer:

8 0

So you don't jump off a cliff thinking you could fly. Laws and Motion are heavily used in jobs and help us understand more about the mechanics of how earth's gravity operates. Imagine in the future where we could manipulate gravity? That's why we learn and study this so much.

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m

Nataly_w [17]

The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

Trial [A] [B] Rate 
 (M) (M) (M/s) 
1 0.50 0.010 3.0×10−3 
2 0.50 0.020 6.0×10−3 
3 1.00 0 .010 1.2×10−2

Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0 =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s

8 0

3 years ago

Read 2 more answers

You have 16.0 g of some compound and you perform an experiment to remove all of the oxygen, 11.2 g of iron is left. What is the

Makovka662 [10]

The empirical formula of this compound is $Fe_2O_3$