Answer:
.
Start color: yellowish-green.
End color: dark purple.
Assumption: no other ion in the solution is colored.
Explanation:
In this reaction, chlorine gas
oxidizes iodine ions
to elemental iodide
. At the same time, the chlorine atoms are converted to chloride ions
.
Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.
Among the reactants of this reaction,
is colorless. If there's no other colored ion, only the yellowish-green hue of
would be visible. Hence the initial color of the reaction would be the yellowish-green color of
.
Similarly, among the products of this reaction,
is colorless. If there's no other colored ion, only the dark purple hue of
would be visible. Hence the initial color of the reaction would be the dark purple color of
.
Answer:Today they will practice measuring different liquids. They will use a container called a graduated cylinder to measure liquids. Graduated cylinders have numbers on the side that help you determine the volume. Volume is measured in units called liters or fractions of liters called milliliters (ml).
Explanation:
Answer:
Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO2). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.
Explanation:
Please give me brainlist
Answer: The pH of the solution is 11.2
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in ml
moles of
=
(1g=1000mg)
Now put all the given values in the formula of molality, we get


pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)

According to stoichiometry,
1 mole of
gives 2 mole of
Thus 0.0298 moles of
gives =
moles of
Putting in the values:
![pOH=-\log[0.0596]=2.82](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B0.0596%5D%3D2.82)



Thus the pH of the solution is 11.2
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