In 1770 a Scottish physician and Chemist Daniel Rutherford performed a simple experiment with which he discovered nitrogen. Rutherford being with an empty bottle that he turned upside down in a pan of water so that the air was trapped. A buring candle was placed inside the bottle with the trapped air causing the water to rise a bit. The part of the air that seemed to "disappear" when the candle was bured was oxygen gas and the part of the air that did not "disappear" Ruthford discovered Nitrogen.
By multiplying the total amount in grams of a given food sample by energy values established by nutritionists for the particular food type i believe
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density 
; we have:
The average atomic weight is:
So; in terms of vanadium, the Concentration of iron is:
From a unit cell volume 
where;
= number of Avogadro constant.
SO; replacing with 
; with 
; 
with 
and
with 
Then:
![a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2FA_%7BFe%7D%20%5D%20%2B%20%5BC_v%2FA_v%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2F%5Crho_%7BFe%7D%20%5D%20%2B%20%5BC_v%2F%5Crho_v%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100-C_v%29A_%7Bv%7D%20%5D%20%2B%20%5BC_v%2FA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100-C_v%29%2F%5Crho_%7Bv%7D%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100A_%7Bv%7D-C_vA_%7Bv%7D%29%20%5D%20%2B%20%5BC_vA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100%5Crho_%7Bv%7D%20-%20C_v%20%5Crho_%7Bv%7D%29%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
Replacing the values; we have:
Answer:
1.99 atm
Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial pressure (P1) = 0.520 atm
Initial temperature (T1) = 26.2°C
Initial volume (V1) = 15.4L
Final temperature (T2) = constant = 26.2°C
Final volume (V2) = 4.02L
Final pressure (P2) =..?
Step 2:
Determination of the new pressure of the gas.
Since the temperature of the gas is constant, it means the gas is obeying Boyle's law. Thus, the new pressure of the gas can be obtained by applying the Boyle's law equation as shown below:
P1V1 = P2V2
0.520 x 15.4 = P2 x 4.02
Divide both side by 4.02
P2 = (0.520 x 15.4) / 4.02
P2 = 1.99 atm
Therefore, the new pressure of the gas is 1.99 atm
Answer:
The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons's law). ... The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law).
