N_2 (g) + 3H_2 (g) rightarrow 2NH_3 (g) volume of H_2 = 32.44 At STP 1 mole of H_2 = 22.4L ? mole of H_2 = 32.4L therefore moles of H_2
Hey there!
Molar mass N2 = 28.01 g/mol
Therefore:
28.01 g N2 -------------- 6.02*10²² molecules N2
( mass N2 ?? ) ----------- 25,000 molecules N2
mass N2 = ( 25,000 * 28.01 ) / ( 6.02*10²³ )
mass N2 = 700250 / 6.02*10²³
mass N2 = 1.163*10⁻¹⁸ g
Hope that helps!
The structure of
Alanine is shown below,
Except the carbon atom of carbonyl group which is
Sp² Hybridized, all remaining atoms are
Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)


Answer:
The answer is below
Explanation:
Because various factors can affect the actual value of the titration outcome. Some of these factors can range from errors from a researcher also known as human error, misreading the quantities, researcher's perception of the color, or wrong procedure in carrying out the experimentation.
Hence, in order to avoid such error, a researcher needs to be thorough during the process of experimentation, and using gross reading can help to avoid these errors when the titre value is eventually determined.