Answer:
The speed of sound in hydrogen gas is 361.5 ![^{\circ} C](https://tex.z-dn.net/?f=%5E%7B%5Ccirc%7D%20C)
Explanation:
It is given that the temperature is 50
.
The speed of sound in meter per second is calculated by an equation.
The equation states that speed of the sound in meter per second is equal to the sum of the product of the temperature in degree celsius with 0.60 and a constant that is 331.5.
The equation is written as:
![speed \ of \ sound \left(m/s \right) = 0.60 \times T + 331.5](https://tex.z-dn.net/?f=speed%20%5C%20of%20%5C%20sound%20%5Cleft%28m%2Fs%20%5Cright%29%20%3D%200.60%20%5Ctimes%20T%20%2B%20331.5)
Put the value of T = 50 then the equation becomes,
![speed \ of \ sound \left(m/s \right) = 0.60 \times 50 + 331.5](https://tex.z-dn.net/?f=speed%20%5C%20of%20%5C%20sound%20%5Cleft%28m%2Fs%20%5Cright%29%20%3D%200.60%20%5Ctimes%2050%20%2B%20331.5)
Speed of sound (in m/s) = 361.5 m/s
Explanation:
![\Delta G^o=-RT\ln K_1](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K_1)
where,
R = Gas constant = ![8.314J/K mol](https://tex.z-dn.net/?f=8.314J%2FK%20mol)
T = temperature = ![600^oC=[273.15+600]K=873.15 K](https://tex.z-dn.net/?f=600%5EoC%3D%5B273.15%2B600%5DK%3D873.15%20K)
= equilibrium constant at 600°C = 0.900
Putting values in above equation, we get:
![\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-%288.314J%2FKmol%29%5Ctimes%20873.15%20K%5Ctimes%20%5Cln%20%280.900%20%29)
![\Delta G^o=764.85 J/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D764.85%20J%2Fmol)
The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.
Equilibrium constant at 600°C = ![K_1=0.900](https://tex.z-dn.net/?f=K_1%3D0.900)
Equilibrium constant at 1000°C = ![K_2=0.396](https://tex.z-dn.net/?f=K_2%3D0.396)
![T_1=[273.15+600]K=873.15 K](https://tex.z-dn.net/?f=T_1%3D%5B273.15%2B600%5DK%3D873.15%20K)
![T_2=[273.15+1000]K=1273.15 K](https://tex.z-dn.net/?f=T_2%3D%5B273.15%2B1000%5DK%3D1273.15%20K)
![\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7B%5CDelta%20H%5Eo%7D%7BR%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
![\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]](https://tex.z-dn.net/?f=%5Cln%20%5Cfrac%7B0.396%7D%7B0.900%7D%3D%5Cfrac%7B%5CDelta%20H%5Eo%7D%7B8.314%20J%2Fmol%20K%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B873.15%20K%7D-%5Cfrac%7B1%7D%7B1273.15%20K%7D%5D)
![\Delta H^o=-18,969.30 J/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D-18%2C969.30%20J%2Fmol)
The ΔH° of the reaction at 600 C is -18,969.30 J/mol.
ΔG° = ΔH° - TΔS°
764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°
ΔS° = -22.60 J/K mol
The ΔS° of the reaction at 600 C is -22.60 J/K mol.
![FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)](https://tex.z-dn.net/?f=FeO%20%28s%29%20%2B%20CO%28g%29%5Crightleftharpoons%20Fe%28s%29%20%2B%20CO_2%28g%29)
Partial pressure of carbon dioxide = ![p_1=P\times \chi_1](https://tex.z-dn.net/?f=p_1%3DP%5Ctimes%20%5Cchi_1)
Partial pressure of carbon monoxide = ![p_2=P\times \chi_2](https://tex.z-dn.net/?f=p_2%3DP%5Ctimes%20%5Cchi_2)
Where
mole fraction of carbon dioxide and carbon monoxide gas.
The expression of
is given by:
![K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_1%7D%7Bp_2%7D%3D%5Cfrac%7BP%5Ctimes%20%5Cchi_1%7D%7BP%5Ctimes%20%5Cchi_2%7D)
![0.900=\frac{\chi_1}{\chi_2}](https://tex.z-dn.net/?f=0.900%3D%5Cfrac%7B%5Cchi_1%7D%7B%5Cchi_2%7D)
![\chi_1=0.900\times \chi_2](https://tex.z-dn.net/?f=%5Cchi_1%3D0.900%5Ctimes%20%5Cchi_2)
![\chi_1+\chi_2=1](https://tex.z-dn.net/?f=%5Cchi_1%2B%5Cchi_2%3D1)
![0.9\chi_2+\chi_2=1](https://tex.z-dn.net/?f=0.9%5Cchi_2%2B%5Cchi_2%3D1)
![1.9\chi_2=1](https://tex.z-dn.net/?f=1.9%5Cchi_2%3D1)
![\chi_2=\frac{1}{1.9}=0.526](https://tex.z-dn.net/?f=%5Cchi_2%3D%5Cfrac%7B1%7D%7B1.9%7D%3D0.526)
![\chi_1=1-\chi_2=1-0.526=0.474](https://tex.z-dn.net/?f=%5Cchi_1%3D1-%5Cchi_2%3D1-0.526%3D0.474)
Mole fraction of carbon dioxide at 600°C is 0.474.
Answer:
<h2>0.17 kg</h2>
Explanation:
The mass of the object can be found by using the formula
![m = \frac{f}{a} \\](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7Bf%7D%7Ba%7D%20%20%5C%5C%20)
f is the force
a is the acceleration
From the question we have
![m = \frac{2}{12} = \frac{1}{6} \\ = 0.166666...](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7B2%7D%7B12%7D%20%20%3D%20%20%5Cfrac%7B1%7D%7B6%7D%20%20%20%5C%5C%20%20%3D%200.166666...)
We have the final answer as
<h3>0.17 kg</h3>
Hope this helps you
Answer:
=> when acid reacts with base
Explanation:
In general, acid-base reactions can be represented by the following ....
acid + base => salt + weak electrolyte
Also, you might read the post at the Brainly link below to better understand this topic
=> brainly.com/question/15984607