Short Answer: <span>Solutions of dilute, weak acids - the only kind that you might taste - are sour. Weak, dilute bases are bitter. Solutions of bases are slippery.
Hope I get brainliest!</span>
Answer:
5.7 × 10⁻³
Explanation:
3.51 - 2.94 = 0.57
Don't change the scientific notation.
0.57 × 10⁻²
BUT
The "0" is not applicable so move the decimal point to the right by one and subtract the power by one.
5.7 × 10⁻²⁻¹ = 5.7 × 10⁻³
The equilibrium reaction that is not shown is the reaction of N₂ and O₂ to form NO:
N₂ + O₂ ⇄ 2NO
We can write an expression for the equilibrium constant. We are given the value of Kc, but we must use Kp if we are dealing with partial pressures. However, since 2 moles of gas are formed from 2 moles of gas, the value of Kc is equal to Kp so we do not need to change anything.
Kc = (Pno)²/(Pn2)(Po2) = 1.71 x 10⁻¹
We must first use the initial pressures to find the reaction quotient to decide which way the equilibrium will shift:
Q = (1.96 x 10⁻³)²/(1.96 x 10⁻³)(1.96 x 10⁻³) = 1
Since Q > Kp, the equilibrium will shift to the left. Now we can prepare an ICE table to find the equilibrium pressures:
N₂ O₂ NO
I 1.96 x 10⁻³ 1.96 x 10⁻³ 1.96 x 10⁻³
C +x +x -2x
E 0.00196 + x 0.00196 + x 0.00196 - 2x
We enter these equilibrium pressures into the equilibrium expression and solve for x:
0.171 = (0.00196 - 2x)²/(0.00196 + x)²
Expand the equation and simplify to a quadratic function:
3.829x² - 0.00851x + 3.183·10⁻⁶ = 0
x = ((0.00851) +/- sqrt((0.00851)² - 4(3.829)(3.183 x 10⁻⁶)))/2(3.829)
x = 0.000476 atm
Now we can solve for the equilibrium partial pressures:
Pno = 0.00196 - 2(0.000476) = 0.00101 atm
Pn2 = 0.00196 + 0.000476 = 0.00244 atm
Po2 = 0.000196 + 0.000476 = 0.00244 atm
Therefore, the equilibrium partial pressure of NO is 0.00101 atm.
