Answer:
The concentration of the most dilute solution is 0.016M.
Explanation:
First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:
<u>First dilution</u>
We can use the dilution rule:
C₁ x V₁ = C₂ x V₂
where
Ci are the concentrations
Vi are the volumes
1 and 2 refer to initial and final state, respectively.
In the first dilution,
C₁ = 0.178 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
<u>Second dilution</u>
C₁ = 0.053 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:
CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol
As observed O2 will not have enthalpy of formation as it is a pure substance.
To calculate for the enthalpy of reaction,
enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
= 294.5 kJ/mol
ANSWER: 294.5 kJ/mol
Answer: Disaccharides
Explanation:
Disaccharides can be defined as the combination of two monosaccharides which combines together to form a disaccharides.
These are formed by the combination of sugars. The process is hydrolysis as it releases water after the reaction is complete.
Two monosachrrides are joined together by the glycosidic linkage. Some common examples are maltose, sucrose and lactose.
Answer:
Volume stays.
Explanation:
If aerosol can throwing into the fire than the temperature of the aerosol can & the temperature of the contents of the gas in an aerosol will also increase, and it induces the pressure to rise.
Against the sides of the can, the gas molecules will smash rapidly with each other which are present inside an aerosol can, and the proportion of gas will stay the same as earlier. According to the law of conservation of matter, as a result of the heat, the gas particles can not be eradicated or expanded.
Answer: 2moles
Explanation:
The best way to solve this problem is to write a balance equation for the reaction.
Which is: Fe2 + 2i2______> 2Fei2
With this you can see that iron ii reacting with excess iodine produces two (2) moles of iron ii iodine.
Alternatively.
Using the formular
Moles = reacting mass/molar mass
Reacting mass = 112g
Molar mass of iron ii = 55g/mol
Hence,
Mole = 112g divides 55g/mol
= 2.0mol//