One separation technique to be used is the paper chromatography. This works by separating the components of the mixture through the difference of their concentrations. There is a stationary phase and the mobile phase, which flows through the stationary phase. The components travel at different rates and is usually signified by the colors. If more than one color would appear, that means that the dye is a mixture.
it will expand as water moves into it.
<span>Answer: option B. 3.07 g
Explanation:
1) given reaction:
S(s) + O₂ (g) → SO(g)
2) Balanced chemical equation:
</span><span>2S(s) + O₂ (g) → 2SO(g)
3) Theoretical mole ratios:
2 mol S : 1 mol O₂ : 2 mol SO
3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa
use the ideal gas equation:
pV = nRT
with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)
=> n = pV / (RT) =
n = [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K) × 573.15 K ]
n = 0.0954 mol SO
4) proportion with the theoretical ratio S / SO
2 mol S x
-------------- = ----------------------
2 mol SO 0.0954 mol SO
=> x = 0.0954 mol S.
5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol
mass = number of moles × atomic mass
mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S
6) Therefore the answer is the option B. 3.07 g
</span>
Answer:
The pH is 7.54
Explanation:
The Henderson - Hasselbalch equation states that for a buffer solution which consists of a weak acid and its conjugate base, the buffer pH is given by:
pH ![=pk_{a} +log(\frac{[conjugate base]}{[weakacid]})](https://tex.z-dn.net/?f=%3Dpk_%7Ba%7D%20%2Blog%28%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bweakacid%5D%7D%29)
pkₐ is for the acid
In this case, the buffer hypochlorous acid HClO is a weak acid, and its conjugate base is the hypochlorite anion ClO⁻ is delivered to the solution via sodium hypochlorite NaClO
.
NaCIO = 0.200 M
HCIO = 0.200 M
pkₐ = -log₁₀ kₐ = -log₁₀ (2.9 × 10⁻⁸) = 7.54
∴pH = 
= 7.54
