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GarryVolchara [31]
3 years ago
8

A student performing this experiment weights out 0.6680 grams of an antacid tablet. It is was determined that this tablet contai

ned 0.4210 grams of CaCO3. What is the percent CaCO3 of this tablet?
Chemistry
1 answer:
Rufina [12.5K]3 years ago
3 0

Answer:

63 wt%

Explanation:

0.4210/0.6680

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c

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8.45 cm - 3.55 cm = ?
lyudmila [28]

Answer:

4.9cm or 4.90cm

Explanation:

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6 0
3 years ago
Read 2 more answers
At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.600 M . N 2 ( g )
alexandr1967 [171]

Answer:

0.78 M

Explanation:

First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:

N2(g) + O2(g) <------> 2NO(g)   Kc = ?

Writting Kc:

Kc = [NO]² / [N2] * [O2]

Replacing the given values we have then:

Kc = (0.6)² / (0.2)*(0.2)

Kc = 9

Now that we have the Kc, let's see what happens next.

We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:

      N2(g) + O2(g) <------> 2NO(g)

I:      0.2        0.2                 0.9

C:     +x         +x                   -2x

E:    0.2+x    0.2+x             0.9-2x

Replacing in the Kc expression we have:

Kc =  [NO]² / [N2] * [O2]

9 = (0.9-2x)² / (0.2+x)*(0.2+x)   ----> (this can be expressed as 0.2+x)²

Here, we solve for x:

9 = (0.9-2x)² / (0.2+x)²

√9 = (0.9-2x) / (0.2+x)

3(0.2+x) = 0.9-2x

0.6 + 3x = 0.9 - 2x

3x + 2x = 0.9 - 0.6

5x = 0.3

x = 0.06 M

This means that the final concentration of NO will be:

[NO] = 0.9 - (2*0.06)

[NO] = 0.78 M

8 0
3 years ago
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