chemistry questions?

OK, my disclaimer is that I had to look up all of these problem types, but I think I may have the right way to do these.

1) You are trying to find the standard heat of formation of ICl. That means that you want to find delta H for the process that makes ICl out of its component parts, at standard temperature and pressure (STP). In other words, if you were walking along at STP and found some iodine and some chloride and put them together to make ICl, what would be the delta H?

To start, define your final equation. At STP, iodine is a diatomic solid: I2(s). At STP, chlorine is a diatomic gas: Cl2(g). Your basic equation, then is to make ICl out of half of a molecule of iodine and half of a molecule of chlorine.

(1/2) I2(s) + (1/2) Cl2(g) -> ICl

Now, in real life you can’t carry out that process in one shot and calculate the heat of formation. So we take advantage of Hess’s Law, which says that you can split this basic process into multiple achieveable steps, find the enthalpy changes for each of those, and then combine them together to get the overall heat of formation.

To do that, rearrange your given equations (you may need to write some of them backwards or multiply a whole equation by a constant). Rearrange them so that if you added all the terms on the left to get your reactants, and added all the terms on the right to get your products, and canceled any terms appearing on both sides, your final equation would be your original target equation. It’s important to remember that if you reverse one of your given equations, you HAVE to change the sign of the delta H. And if you multiply a chemical equation by some constant, you have to multiply the delta H by the same constant. Once those changes are made, add up the enthalpy changes to get the overall heat of formation for ICl. For a good example of how to use Hess’s Law, check the first source I will list.

2) As far as I know, you calculate this by using delta H = (the sum of the enthalpies for products) – (the sum of the enthalpies for the reactants). These values should be in the back of your textbook. Again, make sure that you multiply enthalpies by the number of molecules represented (e.g., in your calculation, use 12 times the enthalpy of water since there are 12 water molecules in the problem). Check the first listed source for more help.

3) If you assume that the container is not absorbing any of the heat (as the problem states), then every bit of energy lost by the copper as it cools down is gained by the surrounding water as it heats up. Heat energy (q) is equal to mass*specific heat*change in temperature, where change in temperature is ALWAYS (final temperature – initial temperature). As the copper drops its temperature from 82.4 degrees to 26.0 degrees, it releases a set amount of heat [qCu = massCu * specific heatCu * (final temperature – initial temperature)]. All of that heat is transferred to the water, so that same q value must ALSO equal qH2O = massH2O * specific heatH2O * (final temperature – initial temperature). Since heat is leaving in one case and arriving in the other, the sign of qCu will be the opposite of qH2O, but their absolute values will be the same. Given that, you can basically set the two equations equal to one another and solve for your only unknown, the mass of the water. Remember that the specific heat of water (which is probably listed in your textbook somewhere) is 4.184 J/°C•g. Check the second listed source for more help.

a million. From a side view of a e book, you will see that there is two facets of the hide and pages. this suggests that the thickness could be: Thickness = 200 * internet site length + 2 * hide length = (200*0.10mm)+(2*0.020mm) = 20.04 mm = 2.004 cm 2. quantity = length * width * top = 10cmx2cmx2xm = 40 cm^3 A 2cmx2cmx2cm block is of quantity 8 cm^3. in case you’re taking the entire quantity of the steel block and divide it with the help of the quantity of each and every individual block, you may discover what share blocks you prefer: 40 cm^3/8cm^3 = 5 blocks Chemistry questions: – Evaporation is the unquestionably approach of taking a substance and changing is from liquid area to gas area. Boiling is the component the place the rigidity exerted with the help of the liquid suits the encompassing rigidity, subsequently inflicting the substance to start evaporating. – the better rigidity you have, the tighter the air is packed, and the greater the flexibility of the molecules is contained and conserved. The power of the molecules, at our point is indicative of warmth. subsequently, the greater rigidity exists, the warmer the contained product will become… which facilitates swifter and greater handy cooking (on the grounds which you may carry onto greater warmth for an prolonged volume of time).