Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
Oxygen because it is on the left of the periodic table so it has a strong pull.
