In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you
Answer:
1.53 × 10²² atoms Ag
Explanation:
Step 1: Define conversions
3.271 × 10⁻²² g = 1 atom
Step 2: Use Dimensional Analysis
= 1.52858 × 10²² atoms Ag
Step 3: Simplify
We have 3 sig figs.
1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag
Answer:
The answer is a. proteins
Answer:
0.1357 M
Explanation:
(a) The balanced reaction is shown below as:
(b) Moles of 
can be calculated as:
Or,
Given :
For :
Molarity = 0.1450 M
Volume = 10.00 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 10×10⁻³ L
Thus, moles of :
Moles of = 0.00145 moles
From the reaction,
1 mole of react with 2 moles of NaOH
0.00145 mole of react with 2*0.00145 mole of NaOH
Moles of NaOH = 0.0029 moles
Volume = 21.37 mL = 21.37×10⁻³ L
Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M
