If you repeat an experiment and the result are very different from the result you got the first time the next step would be to w
1 answer:
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Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:
From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:
= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So,
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
Answer: The mass of the sample will be 1417.7 grams.
Explanation:
We are given:
This means that 1 mole of NaCl has an enthalpy of fusion of 30.2 kJ
1 mole of NaCl has a mass of 58.44 grams.
So, 30.2 kJ of heat is require for a mass 58.44 grams of NaCl
So, 732.6 kJ of heat will be required for = = 1417.65 grams of NaCl.
Hence, the mass of NaCl sample will be 1417.7 grams.
Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law: where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means and
Lastly, we must calculate the number of moles of there are. Given 0.80g of
, we will need to convert with the molar mass of
. Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:
Thus,
Isolating V in the Ideal Gas Law:
...substituting the known values, and simplifying...
So, 0.80g of would occupy 0.56L at STP.