Answer:
0.07172 L = 7.172 mL.
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 = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
Answer:
2.12 moles of gas were added.
Explanation:
We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:
Where in this case:
We <u>input the data</u>:
- 6.13 L * n₂ = 11.3 L * 2.51 mol
As <em>4.63 moles is the final number of moles</em>, the number of moles added is:
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
Answer:
Its B
Explanation:
Products <=> reactants; products can go back to reactant and reactants can form products, depending on the equilibrium position.
The equilibrium position, if it is on the left, and much of the product is formed. If the equilibrium position is not the right, then its the vice versa The equilibrium position is determined by ;
- <em><u>Temperature</u></em>
If the reaction is endothermic, it will be favoured by increase in temperature and equilibrium position will shift to the right ( reactants )
If the reaction is exothermic, its the vice versa
<em><u>N</u></em><em><u>O</u></em><em><u>T</u></em><em><u>E</u></em><em><u>:</u></em><em><u> </u></em>Only temperature affects the equilibrium position
prince Henry is your answer (:
