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Question 8 (2.5 points) Which of the following sets of data are consistent with the law of conservation of matter?7.5 8 of hydrogen gas reacts with 50.0 g oxygen ga...

Question

Question 8 (2.5 points) Which of the following sets of data are consistent with the law of conservation of matter?7.5 8 of hydrogen gas reacts with 50.0 g oxygen gas to form 67.5 & of water: 50 8 gasoline reacts with 243 g oxygen to form 206 g of carbon dioxide and 97 g water 17.7 & nitrogen react with 34.7 g oxygen to form 52.4 & nitrogen dioxide: all of thesenone of these

Question 8 (2.5 points) Which of the following sets of data are consistent with the law of conservation of matter? 7.5 8 of hydrogen gas reacts with 50.0 g oxygen gas to form 67.5 & of water: 50 8 gasoline reacts with 243 g oxygen to form 206 g of carbon dioxide and 97 g water 17.7 & nitrogen react with 34.7 g oxygen to form 52.4 & nitrogen dioxide: all of these none of these



Answers

Are these data sets on chemical changes consistent with the law of conservation of mass? (a) A 7.5 -g sample of hydrogen gas completely reacts with 60.0 g of oxygen gas to form $67.5 \mathrm{g}$ of water. (b) A 60.5-g sample of gasoline completely reacts with 243 g of oxygen to form $206 \mathrm{g}$ of carbon dioxide and $88 \mathrm{g}$ of water.

In order for the law of Conservation of Mass to be followed than the sum of the reactant massive must equal the sum of the product masses for the first reaction the 12.8 and the 19.6 some up to 32.4. And that's the massive product we get for the next one. We have 8 g reacting with 32 g for total 40 g of reactant producing only 17 and 16. That's 33 g of product. So this is not an example of the law of conservation of Mass. Second one is not first one is.

Standard and help ease the formation are only zero for elements, pure elements in their standard state, 25 degrees Celsius and one atmosphere of pressure. Therefore, in looking at all of these chemical reactions, we see that let 1000 degrees is not at 25 degrees Celsius. Although it is an element. C three H eight would not have a delta H of formation equal to zero because it is not an element for part. C. Glucose is not an element, however. For Part D. Nitrogen is an element 25 degrees Celsius and one atmosphere of pressure. It will have a Delta H formation as zero, therefore on Lee de Nitrogen at 25 degrees Celsius, and one atmosphere of pressure would have an intel via formation equal to zero.

Problem, after proto chemical kinetics and equilibrium. So in this ocean, identify the common property for a chemical reaction at dynamic equilibrium. So now let us understand first dynamic equilibrium. So the rate of reaction at any instant is proportional to the concentration of reactant. Now, as the reaction starts as the start of reaction, the concentration of reactant is maximum, and so the rate of forward reaction is maximum. When the chemical reaction is in progress, the concentration of reactant degrees and hence spirit of forward reaction also degrees. At the same time during course of reaction the concentration of product increases, thereby increasing rate of backward reactions. So now let us understand the characteristics of dynamic equilibrium. So the equilibrium is a dynamic process. This state of equilibrium is not affected by the presence of catalyst, it only helps to attain the equilibrium state in less or more time. The observable properties such as a concentration density, color pressure, etcetera, remained constant at constant temperature equilibrium can be attained from either direction. Use of catalyst does not alter a position of a clip them and the forward and backward reaction take this with the same rate, so therefore, from the given options, all the all of the other options are correct so I can write all all the above, which means option day. So hence, therefore, optionally all of the above is correct.

This question wants us to identify the correct statements about octane. So, um, useful information to know before going into this question is that octane has a molar mass of 114.332 grams per mole. And we found that Just buy something. The weights of each Adam, which make up, Come in. And we know the carbon has a move atomic weight of 12.11 grams per mole and that hydrogen has an atomic weight of 1.8 grams per mole. So this first statement, um, 57.1 grams is equal to 0.5 moles of octane. And if you do the math and divide this weight in grams by our molar mass of octane, we should find that it is indeed approximately equal 2.5 moles. So this first statement is correct. Um, next, the question is saying that the compound is 84.1% carbon by weight. And, um, you can double check this by remembering that there are eight carbon atoms and when a molecule octane so then our percent by mass carbon can be calculated by multiplying the atomic weight of carbon by eight Dividing by the massive one mole of octane multiplying the result by 100 you should find it is indeed equal to 84.1 percent. Um, so this second statement is also true. Um, next is the empirical formula of the compound is found by see for age three. Um, this is incorrect. The correct empirical formula is C for H nine. Um, and next question stating that 57.1 grams of octane contains 28 grams of hydrogen. Um, and you can verify this by determining the percent weight of hydrogen. And since he had, you know, the percent weight of carbon, you can just take the difference room 100%. So if the compound is 84.1% carbon by weight than it must be 15 point, um nine, approximately 15.9% hydrogen by weight. And if you multiply 28 grams by that percentage, she'd find it equals to 9.69 grams of Haijun. So this statement this last statement is incorrect.


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