The mass of an unidentified rock is 40 grams. Students determine the volume of the rock by placing the rock in a cylinder with w
ater. The students calculate the density of the rock. They determine the identity of the rock based on the density ranges in the table.
1 answer:
Answer:
A. Coal
Explanation:
= water displaced by rock
Water displaced by rock = volume of water after rock is dropped into the cylinder - volume of water before the rock was dropped into the water
Water displaced by the rock = 180 ml - 150 ml = 30 ml
Density of rock:
40 grams => 30 ml
x grams => 1 ml
Cross multiply
1*40 = 30*x
40 = 30x
40/30 = 30x/30
1.3 = x
Density of rock = 1.3 g per 1 ml
Recall: 1 ml = 1 cm³
Therefore,
Density of the rock = 1.3 g/cm³
1.3 g/cm³ falls within the range of 1.1 - 1.4 g/cm³
Therefore, the rock is identified as Coal.
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Using Hess's law we found:
1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)
2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).
The reactions of dissolution (10.1) and neutralization (10.2) are:
KOH(s) → KOH(aq) ΔHsoln (10.1)
KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)
1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.
Hence, to get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1
KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)
<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:
The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:
Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.
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I hope it helps you!