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vaieri [72.5K]
3 years ago
6

In lab, you calculate the density of an iron rod to be 7.30 g/cm3. The accepted value

Chemistry
1 answer:
OLga [1]3 years ago
6 0

Answer:

<h2>6.41 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 7.80 g/cm³

error = 7.30 - 7.80 = 0.5

We have

p(\%) =  \frac{0.5}{7.8}  \times 100 \\  = 6.410256...

We have the final answer as

<h3>6.41 %</h3>

Hope this helps you

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A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

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So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

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