Answer:
ΔG <0 , ΔH > 0 , ΔS > 0 .
Explanation:
From the data given in question , the reaction is a spontaneous process , hence , the value of change in gibbs free energy would be negative , ΔG <0
And , on dissolution process , the temperature of the water decreases , i.e. , it is an endothermic process , i.e. , the change in enthalphy value is positive , ΔH > 0
And , during the process of dissolution , the ammonia salt break does to ions , i.e. , the randomness increases , hence the ΔS > 0
Answer:
Explanation:
sp² hybridization is found in those compounds having double bond .
Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .
Rest have sp³ hybridization because they are saturated compounds .
Answer:
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
Examples include carbonated water (i.e. soda water); honey; sugar syrup (used in confectionery); supersaturated drug delivery systems. "SDDS"; and sodium acetate solutions prepared from 160 g NaOAc and 30 mL water.
Answer:
0.83 mL
Explanation:
Given data
- Initial concentration (C₁): 12 M
- Final concentration (C₂): 1.0 M
- Final volume (V₂): 10.0 mL
We can calculate the initial volume of HCl using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.0 M × 10.0 mL / 12 M
V₁ = 0.83 mL
The required volume of the initial solution is 0.83 mL.
