Answer:
the answer is C) the iron will repel the substance because it has more electrons than protons
Answer:
Molar mass = 0.09 × 10⁴ g/mol
Explanation:
Given data:
Mass = 0.582 g
Volume = 21.3 mL
Temperature = 100°C
Pressure = 754 mmHg
Molar mass = ?
Solution:
(21.3 /1000 = 0.0213 L)
(100+273= 373 K)
(754/760 = 0.99 atm)
PV = nRT
n = PV/RT
n = 0.99 atm × 0.0213 L / 0.0821 atm. L. mol⁻¹. k⁻¹ × 373 K
n =0.02 mol/ 30.6
n = 6.5 × 10⁻⁴ mol
Molar mass = Mass/ number of moles
Molar mass = 0.582 g / 6.5 × 10⁻⁴ mol
Molar mass = 0.09 × 10⁴ g/mol
The answer
the general for of the equation of reaction is
N2 + 3 H2 ------------ 2NH3
1mol 3mole 2 mole
nH2 nNH3
(nX means number of mole of X)
according to the law of proportion
3 / nH2 = 2 / nNH3 (E)
nNH3 = m NH3 / 17.03 , and m NH3 = <span>0.575 g
consequently </span>nNH3 = 0.575 g / 17.03 = 0.03 mol
by considering (E) 3 / nH2 = 2 / 0.03 = 66.66, and nH2 = 3 / 66.66 =0.045
finally we can write
mH2 = nH2 x 2.0158 =0.045 x 2.0158 = 0.09 g
Answer: Option (d) is the correct answer.
Explanation:
The given reaction will be as follows.

In the ionic form, the equation will be as follows.
........ (1)
............ (2)
Hence, for the net ionic equation we need to add both equation (1) and (2). Therefore, the net ionic equation will be as follows.
Now, balancing the atoms on both the sides we get the net ionic equation as follows.
Answer:
![[A]_{eq}=0.11M](https://tex.z-dn.net/?f=%5BA%5D_%7Beq%7D%3D0.11M)
![[B]_{eq}=0.21M](https://tex.z-dn.net/?f=%5BB%5D_%7Beq%7D%3D0.21M)
![[C]_{eq}=0.19M](https://tex.z-dn.net/?f=%5BC%5D_%7Beq%7D%3D0.19M)
Explanation:
Hello,
In this case, for the given reaction, based on the information about its Gibbs free energy, we obtain the equilibrium constant as shown below:
![Kc=exp(-\frac{\Delta _RG }{RT} )=exp[-\frac{-5240J/mol }{(8.314J/mol*K)(298.15K)} ]=8.28](https://tex.z-dn.net/?f=Kc%3Dexp%28-%5Cfrac%7B%5CDelta%20_RG%20%7D%7BRT%7D%20%29%3Dexp%5B-%5Cfrac%7B-5240J%2Fmol%20%7D%7B%288.314J%2Fmol%2AK%29%28298.15K%29%7D%20%5D%3D8.28)
Now, by means of the law of mass action in terms of the undergoing change
due to the chemical reaction, we obtain:

For which the solution for
by solver is:

Thus, the equilibrium concentrations result:
![[A]_{eq}=0.3M-0.19M=0.11M](https://tex.z-dn.net/?f=%5BA%5D_%7Beq%7D%3D0.3M-0.19M%3D0.11M)
![[B]_{eq}=0.4M-0.19M=0.21M](https://tex.z-dn.net/?f=%5BB%5D_%7Beq%7D%3D0.4M-0.19M%3D0.21M)
![[C]_{eq}=0.19M](https://tex.z-dn.net/?f=%5BC%5D_%7Beq%7D%3D0.19M)
Best regards.