After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Answer:
Final temperature = 83.1 °C
Explanation:
Given data:
Mass of concrete = 25 g
Specific heat capacity = 0.210 cal/g. °C
Initial temperature = 25°C
Calories gain = 305 cal
Final temperature = ?
Solution:
Q = m. c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C
305 cal = 5.25cal/°C × T2 - 25°C
305 cal / 5.25cal/°C = T2 - 25°C
58.1 °C = T2 - 25°C
T2 = 58.1 °C + 25°C
T2 = 83.1 °C
Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
* 
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
* 
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be:
1. related, not
2. how fast molecules vibrate
3. how fast molecules vibrate, the number of molecules
4. higher temperature
5. lower temperature
6. more heat
7. less heat
8. degrees
9. joules
10. expand
1. true
2. false
3. false
4. true
5. false
6. false
7. false
8. true
9. false
10. true
Pure, crystalline solids have a characteristic melting point, the temperature at which the solid melts to become a liquid. The transition between the solid and the liquid is so sharp for small samples of a pure substance that melting points can be measured to 0.1oC. The melting point of solid oxygen, for example, is -218.4o<span>C.</span>
