Answer:
HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)
HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)
HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)
HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)
Explanation:
Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.
- When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).
- When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).
Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.
- When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).
- When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).
Answer: 316.8 g CrSO3
Explanation: Solution:
2.4 moles CrSO3 x 132 g CrSO3 / 1 mole CrSO3 = 316.8 g CrSO4
The conversion factor is 1 mole of CrSO4 is equal to its molar mass which is 132 g CrSO3
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction 
.
![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Answer:
At 430.34 K the reaction will be at equilibrium, at T > 430.34 the
reaction will be spontaneous, and at T < 430.4K the reaction will not
occur spontaneously.
Explanation:
1) Variables:
G = Gibbs energy
H = enthalpy
S = entropy
2) Formula (definition)
G = H + TS
=> ΔG = ΔH - TΔS
3) conditions
ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction
4) Assuming the data given correspond to ΔH and ΔS
ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K
=> T = [ΔH - ΔG] / ΔS
ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K
This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.
The sugar reacts with the gas, turning it to a semi-solid and sticky substance; clogging the gas lines along with many other things.
