Answer:
Original temperature (T1) = - 37.16°C
Explanation:
Given:
Gas pressure (P1) = 2.75 bar
Temperature (T2) = - 20°C
Gas pressure (P2) = 1.48 bar
Find:
Original temperature (T1)
Computation:
Using Gay-Lussac's Law
⇒ P1 / T1 = P2 / T2
⇒ 2.75 / T1 = 1.48 / (-20)
⇒ T1 = (2.75)(-20) / 1.48
⇒ T1 = -55 / 1.48
⇒ T1 = - 37.16°C
Original temperature (T1) = - 37.16°C
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Distance and time is needed for determining speed
Answer:
[H⁺] = 1.0 x 10⁻¹² M.
Explanation:
∵ [H⁺][OH⁻] = 10⁻¹⁴.
[OH⁻] = 1 x 10⁻² mol/L.
∴ [H⁺] = 10⁻¹⁴/[OH⁻] = (10⁻¹⁴)/(1 x 10⁻² mol/L) = 1.0 x 10⁻¹² M.
∵ pH = - log[H⁺] = - log(1.0 x 10⁻¹² M) = 12.0.
∴ The solution is basic, since pH id higher than 7 and also the [OH⁻] > [H⁺].
I think- IDK