Answer:
0.0010m SO₄²⁻
Explanation:
The freezing point depression due the addition of a solute into a pure solvent follows the equation:
ΔT = Kf×m×i (1)
<em>Where ΔT are °C that freezing point decreases (273.15K - 272.47K = 0.68K = 0.68°C). Kf is the constant of freezing point depression (1.86°C/m), m is molality of the solution (0.1778m) and i is Van't Hoff factor.</em>
Van't Hoff factor could be understood as in how many one mole of the solute (sulfuric acid, H₂SO₄), is dissociated.
H₂SO₄ dissociates as follows:
H₂SO₄ → HSO₄⁻ + H⁺
HSO₄⁻ ⇄ SO₄²⁻ + H⁺
<em>Not all HSO₄⁻ dissociates.</em>
1 Mole of H₂SO₄ dissociates in 1 mole of H⁺+ 1 mole of HSO₄⁻ + X moles of SO₄²⁻= 2 + X
Replacing in (1):
0.68°C = 1.86°C/m×0.1778m×i
2.056 = i
Moles of SO₄²⁻ are 2.056 - 2 = 0.056moles SO₄²⁻.
If 1 mole has a concentration of 0.1778m, 0.056moles are:
0.056moles ₓ (0.1778m / 1mole) =
<h3>0.0010m SO₄²⁻</h3>
Answer:
0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM
Explanation:
An enzyme solution is saturated when all the active sites of the enzyme molecule are full. When an enzyme solution is saturated, the reaction is occurring at the maximum rate.
From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.
The addition of more substrate to a solution that contains the enzyme required for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.
<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of 0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>
Answer: C, 36,450.06 moles
Explanation:
Molar mass=478.41
17.19gx478.41moles=36,450.06
2.1 Electrons, Protons, Neutrons, and Atoms
Hydrogen H 1
Helium He 2
Lithium Li 3
Beryllium Be 4
Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: 
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
![AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)](https://tex.z-dn.net/?f=AgCl%20%28s%29%20%2B%202%20NH_3%20%28aq%29%20%2B%20H_2O%20%28l%29%5Crightarrow%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%20%2B%20HCl%20%28aq%29)
; Silver hydroxide at higher temperatures decomposes into black silver oxide: ![2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)](https://tex.z-dn.net/?f=2%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%5Crightarrow%20Ag_2O%20%28s%29%20%2B%20H_2O%20%28l%29%20%2B%204%20NH_3%20%28g%29)
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate: 
. The latter precipitate is yellow.
