Answer: 0.294 mol of 
present in the reaction vessel.
Explanation:
Initial moles of 
= 0.682 mole
Initial moles of = 0.440 mole
Volume of container = 2.00 L
Initial concentration of 
Initial concentration of 
equilibrium concentration of 
The given balanced equilibrium reaction is,
Initial conc. 0.341 M 0.220 M 0 M
At eqm. conc. (0.341-x) M (0.220-x) M (2x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[HBr]^2}{[Br_2]\times [H_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHBr%5D%5E2%7D%7B%5BBr_2%5D%5Ctimes%20%5BH_2%5D%7D)
we are given : (0.341-x) = 0.268 M
x= 0.073 M
Thus equilibrium concentration of = (0.220-x) M = (0.220-0.073) M = 0.147 M
![[Br_2]=\frac{moles}{volume}\\0.147=\frac{xmole}{2.00L}\\\\x=0.294 mole](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%7D%5C%5C0.147%3D%5Cfrac%7Bxmole%7D%7B2.00L%7D%5C%5C%5C%5Cx%3D0.294%20mole)
Thus there are 0.294 mol of present in the reaction vessel.
The person who discovered universal gravitation was newton.
<span>By asking questions and collecting, analyzing, and interpreting data to answer them</span>
Answer:
Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol
Option C is correct
Explanation:
Step 1: Data give
Mass of Ne = 2.0 grams
Molar mass of neon = 20.18 g/mol
Number of Avogadro = 6.0 *10^23 /mol
Step 2: Calculate number of moles of neon
Moles Ne = Mass of ne / Molar mass of ne
Moles Ne = 2.0 / 20.18 g/mol
Moles Ne = 0.099 moles
Step 3: Calculate nulber of particles
Number of particles = 6.022*10^23 / mol * 0.099 moles = 5.96 *10^22
Number of particles = 6.022*10^23 * (2.0g/ 20.18g/mol)
Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol
Option C is correct
Answer is: pH of hydroxylamine solution is 9,23.
Kb(NH₂OH) = 1,8·10⁻⁵<span>.
c</span>₀(NH₂OH)<span> = 0,0500 M =
0,05 mol/L.
c(NH</span>₂⁺) = c(OH⁻) = x.
c(NH₂OH<span>) = 0,05 mol/L - x.
Kb = c(NH</span>₂⁺) · c(OH⁻) / c(NH₂OH).
0,0000000066 = x² / (0,05 mol/L - x).
solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.<span>
pOH = -log(</span>0,000018 mol/L) = 4,74.<span>
pH = 14 - 4,74 = 9,23.</span>
