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tamaranim1 [39]
2 years ago
9

In the lab, you looked at speed-time graphs to determine the acceleration of the cart for each of the three fan speeds. what was

the acceleration of the cart with low fan speed? cm/s2 what was the acceleration of the cart with medium fan speed? cm/s2 what was the acceleration of the cart with high fan speed? cm/s2
Chemistry
1 answer:
lord [1]2 years ago
8 0

Answer:

See below.

Explanation:

Acceleration is the change in velocity over a period of time.  

Let vi be the initial velocity.

Let vf be the final velocity.

t is the time it took to go from vi to vf.

Acceleration is (vf-vi)/t

If the initial velocity was 2 cm/s and the final velocity was 10 cm/s, the object would have incresed it's velocity by (10 - 2) or 8 cm/s.  If iit took 10 seconds to reach vf, the acceleration would be:

(8 cm/s)/(10 s) = 0.8 cm/s^2

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How many molecules are in 4.92 moles of water?
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A 0.20 M solution of a weak acid has a pH of 5.40. What is the Ka for the acid?
sweet [91]

Answer:

The Ka for this weak acid is 7,92 * 10^-11

Explanation:

First of all, let's think the equation

HA + H2O <------> H3O+  +   A-

When we add water to a weak acid, it dissociates in an equilibrium to generate the corresponding anion and the hydronium cation  (the acid form of water)

How do you calculate Ka??  Ka is the acid equilibrium constant.

( [H3O+]  . [A-] ) / [HA] where all the concentrations are in equilibrium.

We don't have the concentration in equilibrium but we have the initial concentration. So...

 HA + H2O <------> H3O+  +   A-

initial- 0.2 M             I don't have H3O+, either A-

reaction - an specific amount reacted  (X)

in equilibrium (0,2 - X)  <----->  X   +  X

And now, how's the formula for Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(X . X) / (0.2-X)

X^2 / (0.2-X) = Ka

Look, that we don't have X as the [H3O+] but we know the pH, so we can know the [H3O+] indeed.

10^-pH = [H3O+]

10^-5,40 = 3,98 * 10^-6

Let's go back to Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(3,98 * 10^-6)^2 / (0.2 - 3,98 * 10^-6) = Ka

(3,98 * 10^-6 is an small number, soooo small that we can approximate to 0)

If we have in order 10^-6, 10^-5 we can consider that.

So now, we have

(3,98 * 10^-6)^2 / (0.2) = Ka

1,58 * 10^-11 / (0.2) = Ka = 7,92* 10^-11

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