Answer:
40.73 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 121.59 kPa/101.325 = 1.2 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 2.0 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas in K (T = 25°C + 273 = 298 K).
<em>∴ V = nRT/P</em> = (2.0 mol)(0.082 L.atm/mol.K)(298 K)/(1.2 atm) = <em>40.73 L.</em>
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.
Raised temperature, decreased volume.
Temperature and Pressure are directly related, when volume increases so does the your pressure.
Volume and Pressure are indirectly related. When volume decreases, your pressure will increase.
Here we assume that disintegration has occur so the normal t½ formular will not be used
mathematically we..⅛×5600 =700
therefore it is about 700years option A
