Answer:
high ph -basic,low pH -acidic
Explanation:
hi I'm from India nice to meet you
A. I think sorry if it’s wrong
Answer:
190 °C
Step-by-step explanation:
The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:
V₁/T₁ = V₂/T₂ Invert both sides of the equation.
T₁/V₁ = T₂/V₂ Multiply each side by V₂
T₂ = T₁ × V₂/V₁
=====
V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K
V₂ = 5.6 L; T₂ = ?
=====
T₂ = 310.15 × 5.6/3.75
T₂ = 310.15 × 1.49
T₂ = 463 K
t₂ = 463 – 273.15
t₂ = 190 °C
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
