Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
To determine the moles of KIO_3 titrated, use the balanced equation 2 KIO_3 + 5 Na_2S_2O_3 + 6 HCl → 3 I_2 + 6 NaCl + 6 NaClO + 3 H_2O. Therefore, 0.001551 mol of KIO_3 were titrated.
To determine the moles of KIO3 titrated, we need to use the balanced equation for the reaction:
2 KIO3 + 5 Na2S2O3 + 6 HCl → 3 I2 + 6 NaCl + 6 NaClO + 3 H2O
From the equation, we can see that 2 moles of KIO3 react with 5 moles of Na2S2O3. Therefore, the moles of KIO3 titrated can be calculated using the following proportion:
(0.0100 M KIO3 / 1 L) * (15.51 mL / 1000 mL) * (2 mol KIO3 / 5 mol Na2S2O3) = 0.001551 mol KIO3
#SPJ12
Answer:
Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane.
Explanation:
A double bond is converted to an oxirane through oxidation by peracids e.g. mCPBA (meta-chloroperoxybenzoic acid).
Epoxidation can occur at both face of double bond result in formation of two stereoisomers.
Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane
Both A and B contain plane of symmetry. Hence, both the products are achiral. So, they do not rotate the plane of polarization of plane polarized light.
phosphate (Ca3(PO4)2), a major component of bone.
% Ca
%P
% 0
Answer:
38.7%
41.3%
20%
Explanation:
The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.
To solve this problem:
Compound:
Ca₃(PO₄)₂
Formula mass = 3(40) + 2[31 + 4(16)]
= 120 + 2(95)
= 120 + 190
= 310
%C = x 100 = 38.7%
%P = x 100 = 41.3%
%O = x 200 = 20%
Fotón
Glúon
Leptón
Answer:
Leptón
Explanation:
Lepton son partículas elementales de espín de medio entero (espín 1⁄2), y se sabe que no experimentan interacciones fuertes. Los leptones se clasifican en leptones cargados (los leptones similares a los electrones) y leptones neutros (conocidos como neutrinos). Un electrón es un leptón. Los leptones cargados pueden combinarse con otras partículas para formar átomos compuestos de partículas y positronio, mientras que los neutrinos rara vez interactúan con algo, lo que los hace raros de observar.
Dado que la electricidad es el resultado del movimiento o flujo de electrones, la única partícula representativa de esta fuerza es el leptón.
Explanation:
There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.
The question is incomplete, complete question is ;
A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.
Answer:
Partial pressure of the oxygen gas is 3.91 atm.
Partial pressure of the helium gas is 20.0 atm
Total pressure of the gases is 24.0 atm
Explanation:
Moles of oxygen gas =
Moles of helium gas =
Total moles of gas =
Volume of the cylinder = V = 10.0 L
Total pressure in the cylinder = P = ?
Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K
PV = nRT ( ideal gas equation )
P = 23.88 atm ≈ 23.9
Partial pressure of the individual gas will be determined by the help of Dalton's law:
partial pressure = Total pressure × mole fraction of gas
Partial pressure of the oxygen gas
Partial pressure of the helium gas