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3 years ago

13

What is the oxidation number of iodine in KL04? (1)- +1 (2)- -1 (3)- +7 (4)- -7

1 answer:

Nostrana [21]3 years ago

4 0

3) +7

I got this stupid question wrong so that you people don't have to. I simply hate chemistry and I wish it didn't exist. Kind of like this website that makes me explain the answer.

Trust me, the answer is correct

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Water's ability to dissolve a wide variety of molecules is important, but more important is the hydrophobic effect, which drives

Sliva [168]

Answer:

d. Hydrophobic molecules are attracted to each other.

Explanation:

The term “hydrophobic effect” is associated with the spontaneous tendency of macromolecules, such as proteins, to prefer a conformation in an aqueous medium, with hydrophobic groups facing the interior of the mac romolecule, favoring attractive intramolecular interactions, and hydrophilic groups exposed on the surface, for maximize interactions with water molecules in the medium. This is because the hydrophobic molecules are attracted to each other, allowing them to turn inward.

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3 years ago

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2 years ago

5. __NH3 + __O2 >>>>>__ NO +__ H20

inn [45]

2 NH3+ 2 O2 —> 2 NO+ 3 H2O

5 0

3 years ago

What do the atoms of a metal look like when they are in a body-centered cubic arrangement? A a cube with a ball stuck on each of

Natali [406]

<span>B) a cube with a ball stuck on each of its eight corners and one suspended at its center </span>

4 0

3 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago