Which element is the most reactive?

What is the molarity of a solution of h3po4 if 10.2 ml is neutralized by 53.5 ml of 0.20m koh what is the molarity of a solution

Anna35 [415]

The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M

<h3>Balanced equation </h3>

H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O

From the balanced equation above,

The mole ratio of the acid, H₃PO₄ (nA) = 1
The mole ratio of the base, KOH (nB) = 3

<h3>How to determine the molarity of H₃PO₄ </h3>

Volume of acid, H₃PO₄ (Va) = 10.2 mL
Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
Volume of base, Ca(OH)₂ (Vb) = 53.5 mL

Molarity of acid, H₃PO₄ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 10.2) / (0.2 × 53.5) = 1 / 3

(Ma × 10.2) / 10.7 = 1 / 3

Cross multiply

Ma × 10.2 × 3 = 10.7

Ma × 30.6 = 10.7

Divide both side by 30.6

Ma = 10.7 / 30.6

Ma = 0.35 M

Learn more about titration:

brainly.com/question/14356286

#SPJ1

8 0

3 years ago

Which is NOT an isotope of carbon?

Strike441 [17]

Carbon-6 because it is a neutron number

5 0

3 years ago

Read 2 more answers

What happens to an oxidizing agent during a redox reaction?

Gnoma [55]

Answer:

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. ... A reducing agent is oxidized, because it loses electrons in the redox reaction.

Explanation:

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. ... A reducing agent is oxidized, because it loses electrons in the redox reaction.

4 0

3 years ago

Use the collision theory to explain how increasing the surface area of a solute will increase the rate of the dissolving process

Maksim231197 [3]

<span>More surface area --> more molecules of the solute in contact with the solvent --> more chance for a solvent molecule to collide with the solute molecules --> dissolves faster

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>

5 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

4 years ago