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I am Lyosha [343]
3 years ago
6

43 milliliters of water weighs 43 g. what is the density of the water?

Chemistry
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

\rho =1g/mL

Explanation:

Hello,

In this case, since the density is defined as the ratio between the mass and the volume as shown below:

\rho =\frac{m}{V}

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

\rho =\frac{43g}{43mL}=1g/mL

Regards.

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Answer:

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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

α = 0.103 . 2/3 ⇒ 0.0686

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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