Will give brainliest!!!!

Vika [28.1K]

I believe the medulla is the answer.

3 0

3 years ago

What is redox reaction?? <img src="https://tex.z-dn.net/?f=%20%5C%5C%20%20%5C%5C%20" id="TexFormula1" title=" \\ \\ " alt="

MAXImum [283]

Answer:

Oxidation–reduction or redox reactions are reactions that involve the transfer of electrons between chemical species (check out this article on redox reactions if you want a refresher!). The equations for oxidation-reduction reactions must be balanced for both mass and charge, which can make them challenging to balance by inspection alone. In this article, we’ll learn about the half-reaction method of balancing, a helpful procedure for balancing the equations of redox reactions occurring in aqueous solution.

Explanation:

8 0

3 years ago

Read 2 more answers

The molecular mass of a substance refers to the:

ololo11 [35]

Answer:

A

Explanation:

it's the sum of the masses of all the atoms in the molecule

please like and Mark as brainliest

3 0

3 years ago

A large crystal of potassium manganate is placed in the bottom of a beaker of cold water, and left for several hours. Describe w

Katarina [22]

Potassium manganate would dissolve in water

5 0

2 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

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

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

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago