Answer:
0.3811 mol.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
∵ P = 2.1 atm, V = 4.5 L, T = 302 K, R = 0.0821 L.atm/mol.K.
<em>∴ n = PV/RT </em>= (2.1 atm)(4.5 L)/(0.0821 L.atm/mol.K)(302.0 K) = <em>0.3811 mol.</em>
Answer:
True
Explanation:
Ethers react with HI to form the corresponding alcohols and alkyl iodides.
Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes
reaction with HI to form ethyl iodide.
<u>Hence, ethyl iodide is the only product when ethyl ether reacts with excess of HI for several hours.</u>
Answer:
One gallon of octane produces approximately 7000 L of carbon dioxide.
Note:
I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.
Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.
I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
so the reaction NaHCO3 + HCl --> NaCl + H2CO3
mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3