Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
The third one
synthesis reactions have multiple reactants that synthesize into one product
Reducing acidity of chyme :Acidic chyme entering the duodenum stimulates the release of secretin from the small intestinal glands.
Answer:
Molarity = 0.7 M
Explanation:
Given data:
Volume of KCl = 20 mL ( 0.02 L)
Molarity = 3.5 M
Final volume = 100 mL (0.1 L)
Molarity in 100 mL = ?
Solution:
Molarity = number of moles of solute / volume in litter.
First of all we will determine the number of moles of KCl available.
Number of moles = molarity × volume in litter
Number of moles = 3.5 M × 0.02 L
Number of moles = 0.07 mol
Molarity in 100 mL.
Molarity = number of moles / volume in litter
Molarity = 0.07 mol /0.1 L
Molarity = 0.7 M
The cost of one antacid is 2.325 cents per tablet.
<u>Explanation:</u>
As per the question based on the student analysis we know that,
Total antacid tablets in a bottle = 120
Purchase Price of a bottle = $ 2.79
Cost of 1 antacid tablet 
As we know $1 = 100 cent
The cost of 1 antacid tablet = 
× 100 cents = 2.325 cents/tablet
.
Thus we came to know that it costs 2.325 cents/tablet
.
