Answer:
Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.
Explanation:
Make sure have same amounts of species on both sides
Cu (s) + 2 AgNO3 (aq) -> Cu(NO3)2 (aq) + 2 Ag (s)
Answer:
Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal
Explanation:
The molecular formula of mannoheptulose is C₇H₁₄O₇.
The structure is as shown in the attachment below.
Number of C-C bonds present in mannoheptulose = 6
Number of C-H bonds present in mannoheptulose = 8
Since the each C-C bond contains 76 Kcal of energy,
Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal
Also, since each C-H bond contains 91 Kcal of energy;
amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal
Total Kcal energy produced in the catabolism of mannoheptulose = 456 + 728 = 1184 Kcal
For the answer to the question above, asking to w<span>rite the complete balanced equation for the reaction between aluminum metal (Al) and oxygen gas (O2)and You do not need to make the subscripts smaller.
My answer would be,
</span><span>4Al(s) + 3O2(g) --->2 Al2O3(s)
</span>
I hope this helps.
Sliding or rolling friction
