Answer is: <span>the pressure of the gas is 9,2 atm.
</span>p₁ = 4,0 atm.
T₁ = 300 K.
V₁ = 5,5 L.
p₂ = ?
T₂ = 250 K.
V₂ = 2,0 L.
Use combined gas law - the volume of amount of gas is proportional to the ratio of its Kelvin temperature and its pressure.<span>
</span>p₁V₁/T₁ = p₂V₂/T₂.
4 atm · 5,5 L ÷ 300 K = p₂ · 2,0 L ÷ 250 K.
0,0733 = 0,008p₂.
p₂ = 9,2 atm.
Answer:
ΔHrxn = - 1534.3 J
Explanation:
Given the assumptions and the formula for the change in enthalpy:
ΔHrxn = m x C x ΔT, where
m is the mass of solution given 135.4 g
C is the heat capacity 4.2 J/g .K and,
ΔT is the change in temperature
we have ,
T₁ = ( 18.1 + 273) K = 291.1 K
T₂ = ( 15.4 +273) K = 288.4 K
ΔHrxn = 135.3 g x 4.2 J/gK x ( 288.4 -291.1 ) K = - 1534.3 J
After verifying our result has the correct unit, the answer is -1534.3 Joules, and the negative sign tells us it is an endothermic reaction decreasing the final temperature.
Answer: the ability to be dissolved, especially in water.
Explanation: I think the answer you've picked is right
Hope this helps
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M