Help me please omg Answer the following question:

How many moles of ammonia are in 375. mL of a 2.25 M aqueous ammonia solution?

Find the numerical answer for this question and make sure to include the following:

What is the chemical formula for ammonia?
What is the formula for molarity?
When you give your numerical answer, what is the correct significant figures and how do you know that is the correct amount?

NH₃

M = n/V(L)

0.844 mol (Both numbers have 3 significant figures so the result has 3 significant figures as well)

Explanation:

Step 1: Given and required data

Step 2: Calculate the moles (n) of ammonia (solute)

Molarity is equal to the moles of solute divided by the liters of solution.

M = n/V(L)

n = M × V(L)

n = 2.25 mol/L × 0.375 L = 0.844 mol (Both numbers have 3 significant figures so the result has 3 significant figures as well)

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%