V1 = 445ml V2 = 499ml
T1 = 274 K T2 = ?
By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K
Answer:
Regarding Measures of center, the Mean will be smaller, and Median will not be affected. Thus, the median will not change, since the temperature was already low initially and only decreased. However, the median takes the average of all values into account, and if you substitute a value for a smaller value, the average decreases.
In relation to Measures of spread, the range and standard deviation will be larger, and the IQR will not change. Therefore, the range is the highest minus the lowest number; if the lowest number decreases sharply, the range increases. On the other hand, the IQR only counts the first and third quartile values. Therefore, since the value was already the lowest, if it is decreased, it will not change the IQR. However, the standard deviation will increase significantly, as the decreasing value of the lowest data point will increase the overall spread.
Explanation:
Answer:
2K+ + NO-2 -> K2NO
Explanation:
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Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
D) Chlorine is reduced as hydrogen is added whereas Iron is oxidised as Joins with a non hydrogen compound.
