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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:

Two factors that might have a affect of which copper sulphate mineral will occur at a given location  is:

A. Copper sulphate high solubility in water

B. Also it binds nicely with the sediments  or the crystal.

Explanation:

As it is mentioned here that copper sulphate can be crystallized as an anhydrate which means that their  is no waterin those crystals or can be as of those three different hydrates whose crystal structure varies with the amount of water present in it.

The four forms are also given of the copper sulphate are:

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So, the two factors that might give an affect which type of copper sulphate  mineral willoccur at a given location is:

A. The copper sulphate high solubility in water.

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what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

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ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

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