100 degree celcius and 0 degree Celsius
Explanation:
A process in which water vapor changes into liquid state is known as condensation. As we know that when energy is released in a reaction then it is known as exothermic reaction and when energy is absorbed in a reaction then it is known as endothermic reaction.
As vapors have high energy so, when they change into liquid state then heat energy is released by them. Therefore, condensation is an exothermic reaction.
As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.
So, when there will occur a decrease in temperature then molecules of a gas will come closer to each other. Hence, there will also occur a decrease in vapor pressure of the gas.
A) -0.5(9.8)*t^2 = -25(t-2) - 0.5(9.8)(t-2)^2
-4.9t^2 = -25t + 50 - 4.9(t^2-4t+4)
0 = -25t+50+19.6t - 19.6
5.4t = 30.4
t = 5.62962963 s
b) h = -4.9(5.62962963)^2
h = -155.2943759
the building is 155.2943759 m high
c) speed 0of first stone
= at
= 9.8*5.62962963
= 55.17037037 m/s
speed of second stone
= v + at
= 25+9.8*3.62962963
= 60.57037037 m/s
Answer:
392g sulfuric acid are produced
Explanation:
Based on the balanced equation:
2HCl + Na2SO4 → 2NaCl + H2SO4
<em>2 moles of HCl produce 1 mole of sulfuric acid</em>
<em />
To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -<em>Molar mass H2SO4 = 98g/mol- </em>we can find the mass of sulfuric acid produced:
<em>Moles sulfuric acid:</em>
8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4
<em>Mass sulfuric acid:</em>
4mol H2SO4 * (98g / mol) =
392g sulfuric acid are produced
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):

mass O₂ = 120 g
mol O₂(MW=32 g/mol) :

Mol ratio of reactants(to find limiting reatants) :

mol of H₂O based on O₂ as limiting reactants :
mol H₂O :

mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
