We first have to find Q calorimeter
Q calorimeter = C (delta T)
C= heat capacity in J/oC, not small c ( specific heat capacity)
delta T= change in T (oC) = final T - initial T
Q calorimeter = 10.4 J/oC (30 oC - 25 oC)
Q calorimeter = 52 J
Remember,
Q calorimeter = - Q surrounding
The (-) negative sign stands for heat being released.
So ,
Q surrounding = - 52 J
In other words, the iron releases 52 J .
The calorimeter absorbs the entire 52 J and heat rose from 25 - 30 oC.
Answer: 52 J
The reaction of acid, assuming HCl and calcium carbonate always produces a gas. The reaction is as follows:
2 HCl + CaCO3 --> CaCl2 + H2CO3
H2CO3, carbonic acid, is a weak acid that is unstable in water solutions at high concentrations. As such, it decomposes:
H2CO3 --> H2O + CO2
Then,
2 HCl + CaCO3 --> CaCl2 + H2O + CO2
The total ionic equation looks as follows:
2H+(aq) + 2 Cl-(aq) + CaCO3(s) --> Ca+2(aq) + 2 Cl-(aq) + H2O(l) + CO2(g)
Clearly, Cl- is a spectator ion as it is unchanged in the reaction. The net ionic reaction looks as follows:
2 H+(aq) + CaCO3(s) --> Ca+2(aq) + H2O(l) + CO2(g)
A. Dalton's theory that atoms could not be divided was incorrect
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
