How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

Molarity of the acid (Ma) = 6M

Volume of the acid (Va) =?

Volume of the base (Vb) = 39mL

Molarity of the base (Mb) = 2M

Step 3:

Determination of the volume of the acid.

Using the equation:

MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

6 x Va / 2 x 39 = 1/1

Cross multiply to express in linear form

6 x Va = 2 x 39

Divide both side by 6

Va = (2 x 39)/6

Va = 13mL

Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.