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Arada [10]
3 years ago
11

The table below shows the density of a sample of a mystery liquid you tested in the lab. Can you infer the identity of the subst

ance from these data?
A. Yes, the substance must be water.
B. No, more data are needed.
C. No, the data must be wrong because density always decreases with an increase in temperature.
D. Yes, but only if the data for 50ºC and 70ºC were also present.

Chemistry
1 answer:
Mashutka [201]3 years ago
7 0

Answer:

A. Yes, the substance must be water.

Explanation:

The density of a substance is unique to it. Density is defined the as the amount of substance contained per volume.

One of the ways of identifying a substance is to determine its density. Every matter is known to have their own specific densities. This makes them different from other substances. The density of gold is unique to it and it differs from that of silver.

In fact, water has density of 1.00gcm⁻³. Experimental errors and some little factors must have altered our expected figure. This a case of precision and accuracy in the experiment.

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Answer:

B

Explanation:

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3 years ago
Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
4 years ago
Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3
jeyben [28]
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
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6 0
3 years ago
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Name the following compound
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Answer:

bromine

Explanation:

hoefully this is right

8 0
3 years ago
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