Answer:
26.110 grams of O2 produced.
Explanation:
Calculate the amount of moles in KClO3 by dividing the amount of grams given by the atomic weight of the substance.
To get the atomic weight: K = 39.098, Cl = 35.45, O = 15.999, and there are 3 molecules of Oxygen, so multiply 15.999 by three.
39.098 + 35.45 + (15.999 * 3) = 122.548.
66.7g / 122.548 atomic mass = 0.544 moles.
The ratio of moles of KClO3 to moles of O2 is 2 to 3.
= 
Cross multiply to get 1.632 = 2y. Y = 0.816, meaning 0.816 moles of O2 will be produced.
Convert this into grams by multiplying by the atomic weight of O2 (15.999 * 2 = 31.998).
0.816 * 31.998 = 26.110 grams of O2 produced.
Answer : The 
must be administered.
Solution :
As we are given that a vial containing radioactive selenium-75 has an activity of 
.
As, 3.0 mCi radioactive selenium-75 present in 1 ml
So, 2.6 mCi radioactive selenium-75 present in 
Conversion :
Therefore, the must be administered.
Answer:
En la imagen adjunta.
Explanation:
¡Hola!
En este caso, dados los lineamientos IUPAC que tenemos en cuenta para dibujar la estructura de hidrocarburos cíclicos, para el 1,4-ciclohexadieno, es posible determinar que esta estructura viene dada por hexágono que representa el ciclohexano y dos dobles enlaces en los carbonos 1 y 4 al ser un dieno. Por lo anterior, en la imagen adjunta, puede encontrar la estructura requerida.
¡Saludos!
Answer:
-3
Explanation:
The oxidation state or oxidation number of an atom is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
The complex anion here is [Cr(CN)6]3-.
Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),
so x= +3. Hence the oxidation state of Chromium in this complex hexacyanochromium (III) anion comes out to be -3.
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