Answer:
(a) 133.33nm
(b) 600nm
(c) 7,600nm
Explanation:
The concentration of Y can be determined by using the formula:
where;
[L] = concentration of the binding ligand.
kd = 400 nm
Thus:
When Y = 0.25; we get :
0.25 (400 + [L]) = [L]
100 + 0.25[L] = [L]
100 = [L] - 0.25 [L]
100 = 0.75 [L]
[L] = 100/0.75
[L] = 133.33 nm
At, Y = 0.6
0.6 (400 + [L]) = [L]
240 + 0.6[L] = [L]
240 = [L] - 0.6 [L]
240 = 0.4 [L]
[L] = 240/0.4
[L] = 600 nm
At, Y = 0.95
0.95 (400 + [L]) = [L]
380 + 0.95[L] = [L]
380 = [L] - 0.95 [L]
380 = 0.05 [L]
[L] = 380/0.05
[L] = 7600 nm
Answer:
1026.88 kilocalories are released by the combustion of 1.50 mol of glucose
Explanation:
First we have to calculate the amount in grams in 1.5 moles of glucose by using formula
Molecular mass of glucose is 180.156 g/mol
Mass in grams = Molarity x Molecular mass
=1.50 moles x 180.156 g/mol
= 270.23 g
Now we will calculate the amount of energy released
Energy released by 1 gram of glucose = 3.8 Kcal
Energy released by 270.23 gram of glucose = 3.8 x 270.23 Kcal
= 1026.88 Kcal
1026.88 kilocalories are released by the combustion of 1.50 mol of glucose
100.4 degree celsius is the temperature of the water having non-electrolyte.
The correct answer is b.
Explanation:
Data given:
mass of non-electrolyte solute = 31.65 grams
volume of water = 220 ml or 0.22 litres
molar mass of solute = 180.18 grams/mole
density of water = 1 gm/mole
Boiling point of water = 100 degrees celsius
molality = ?
Kb for water = 0.51
boiling point elevation or temperature of hot water,T =?
Formula used:
T = mKb equation 1
number of moles of non electrolyte =
number of moles = 0.17 moles
molality =
molality = 0.77 M
putting the values in equation 1
T = 0.77 X 0.51
T = 0.392 degree celsius. is the elevation in temperature when solute is added.
Temperature of the hot water = 100 +0.392
= 100.392 degree celsius.
Answer:
B. <u><em>At the end of the reaction, both product and reactants are of a constant concentration.</em></u>
Explanation:
- Option A and C are similar as they both refer to constant concentration of product and reactant respectively in first half. As in the graph, the concentration of reactant and product changes (concentration of reactant decreases and concentration of product increase) with time in the <em>first half</em> of the reaction. This made both A and C option wrong.
- Option D is also wrong as at the <em>end of reaction</em> the line of concentration of product and reactant do not coincide which means they are not equal.
- Option B is correct as we take the end of reaction at the point where the concentration of reactant and product won't change much or become constant
<u><em>first half</em></u> time is the when concentration of reactant reduces to 50% of initial concentration which you can nearly assume on or before the point of intersection of both the concentration graphs.
<u><em>end of reaction</em></u><em> </em>is the time when the reaction completes which is theoretically infinite but generally we take end of the reaction as the time when the slope of concentration curve becomes nearly zero or the time when change in concentration of reactant and product is negligible.
Answer:
The correct answer is 0.165 g NaCl.
Explanation:
The following is the precipitation reaction taking place between sodium chloride and silver nitrate:
NaCl (aq) + AgNO₂ (aq) ⇒ NaNO₃ (aq) + AgCl (s)
The complete ionic reaction of the reaction will be,
Na⁺ + Cl⁻ + Ag⁺ + NO₃⁻ ⇒ AgCl (s) + Na⁺ + NO₃⁻
Hence, the net ionic equation for the mentioned reaction is:
Ag⁺ (aq) + Cl⁻ (aq) ⇒ AgCl (s)
Thus, it can be witnessed that one mole of chloride ion is needed so that one mole of Ag⁺ ion get neutralized. There is a need to find the moles of silver ions present in the solution of AgNO₃ and then transform these moles to the moles of chloride ion. Ultimately, these moles can be converted to the concentration of sodium chloride needed.
The no. of moles of silver ions found in silver nitrate solution is,
(2.50 × 10² mL) × (0.0113 mol Ag⁺/1000 ml solution) = 2.83 × 10⁻³mol Ag⁺
Now the moles of chloride ions needed to precipitate the silver ions is,
(2.83 × 10⁻³ mol Ag⁺ ) × (1 mol Cl⁻/1 mol Ag⁺) = 2.825 × 10⁻³mol Cl⁻
The mass of sodium chloride needed for precipitating the silver ions will be,
mass of NaCl = (2.83 × 10⁻³ mol Cl⁻) × (1 mol NaCl / 1 mol Cl⁻) × (58.44 grams / 1 mol NaCl)
= 0.165 gram NaCl.