Carbon cannot reduce sodium because sodium is stronger than carbon. reactive metals like sodium and potassium and electrolysis has to be used.
Explanation:
High density medium to low density medium.
Answer: The final temperature would be 1250.7 K.
Explanation: We are given a sample of helium gas, the initial conditions are:
(Conversion factor: 1L = 1000 mL)
(Conversion Factor: 1° C = 273 K)
The same gas is expanded at constant pressure, so the final conditions are:
To calculate the final temperature, we use Charles law, which states that the volume of the gas is directly proportional to the temperature at constant pressure.
Putting the values, in above equation, we get:
To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16
Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.
C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96
Then add them up.
72.06+ 12+ 96= 180.06
Now find the percent composition of carbon.
72.06/ 180.06 x 100= 40.01%
So the answer is C 40%.
The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3
