A student has 475 mL of a 4.5 M HNO3 solution. He leaves the top of the container Monday morning the volume of solution is now 347 mL. What is the conventtration of the solution now

6.1 M

Explanation:

Step 1: Calculate the moles of HNO₃ in the initial solution

A student has 475 mL (0.475 L) of a 4.5 M HNO₃ (4.5 mol HNO₃/ L) solution.

0.475 L × 4.5 mol HNO₃/ L = 2.1 mol HNO₃

He leaves the top of the container open, so water will evaporate, but the moles of HNO₃ will remain the same.

Step 2: Calculate the final concentration of the solution

2.1 moles of HNO₃ are in 347 mL (0.347 L) of solution.

[HNO₃] = 2.1 mol / 0.347 L = 6.1 M

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.