Which branch of chemistry would study how pollution effects marine life?

A student analyzed an unknown sample that contained a single anion. The sample gave a yellow precipitated upon addition of a sol

kenny6666 [7]

<h3>Answer:</h3>

Anion present- Iodide ion (I⁻)

Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)

<h3>Explanation:</h3>

In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.

Additionally we need to know the color of the precipitates.

Some of insoluble salts of silver and their color include;

Silver chloride (AgCl) - white color
Silver bromide (AgBr)- Pale cream color
Silver Iodide (AgI) - Yellow color
Silver hydroxide (Ag(OH)- Brown color

With that information we can identify the precipitate of silver formed and identify the anion present in the sample.

The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
Therefore, the anion that was present in the sample was iodide ion (I⁻).
Thus, the corresponding net ionic equation will be;

Ag⁺(aq) + I⁻(aq) → AgI(s)

4 0

3 years ago

1. what part (or parts) of the system store potential energy?

AysviL [449]

Answer:

Khud karo warna teacher ko bata don ga

Explanation:

8 0

3 years ago

Important functional groups in biomolecules include --------

Andrew [12]

B nucleic acids I think

4 0

2 years ago

Read 2 more answers

If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above

jasenka [17]

Answer : The density of an object is, $2g/cm^3$

Solution : Given,

Mass of an object = 60 g

Volume of an object = $30cm^3$

Formula used :

$\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}$

Now put all the given values in this formula, we get the density of an object.

$\text{Density of an object}=\frac{60g}{30cm^3}=2g/cm^3$

Therefore, the density of an object is, $2g/cm^3$

6 0

2 years ago

0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium

Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002$

From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

$\tt \dfrac{2}{5}\times 0.002=0.0008$

6 0

3 years ago