Intermediate is created and destroyed, and therefore does not appear on the net equation, which is . Thus, the intermediate is . Note that when asked for an intermediate, the coefficient in front of it is not used, rather we are looking for the species that is a product of one reaction and a reactant in a subsequent step.

The reactant  is not included in the rate law expression, and therefore altering its concentration does not affect the rate of the reaction. Catalysts always increase the rate of reactions by lowering its activation energy. Increasing temperature (average kinetic energy of the molecules) increases the frequency of collisions, and increases the proportion of collisions that have enough energy to overcome the activation energy and undergo a chemical reaction. Increasing the concentration of  will increase the rate of the reaction as indicated by the rate law.

For this question, we're asked to identify something that will always increase the rate of a reaction. Notice that the question specifically states "always."

Let's go through each answer choice and see how it affects the reaction rate.

When the concentration of reactants is increased, this may increase the reaction rate, but not always. For example, if a reaction is zero-order with respect to its reactants, then changing the reactant concentration will have no effect on the rate.

Just like with reactants, decreasing the concentration of products may or may not change the reaction rate. If the reaction rate is zero-order with respect to the products, then a change in their concentration will have no effect on the reaction rate.

A change in temperature is the only thing that is guaranteed to change the reaction rate. This is because changing the temperature will directly change the rate constant of the reaction. Increasing the temperature will increase the rate constant, and hence the reaction rate.

We need to find the specific heat of the unknown sample of metal in order to locate it on the list. We can do this by using the equation that allows us to determine the specific heat capacity of an element.

Since we know the change in temperature, we can simply plug in the values and solve for the value of .

Going back to the list, we see that this is the specific heat capacity for copper, so we confirm that the unknown metal is copper. 

Bomb calorimeters are most useful when dealing with a gas, because they can operate well at high pressures. Coffee-cup calorimeters are not useful when water begins to boil, producing vapor.

You need heat for the phase change, using the enthalpy of fusion (100g*334 J/g = 33400 J). Add to this the heat to get to boiling point using the specific heat of water (100g*100C*4.2 J/g^oC = 42000 J). Totalling 75400 J (75.4 kJ)

First we will determine the mass of the liquid:

Now we will examine the relationship between heat and specific heat capacity:

Where is heat in Joules, is the specific heat capacity,  is the mass and is the change in temperature. We can rearrange this

If we begin at , we will end at

When the heated metal is placed in the container of the cooler water there will be a transfer of thermal energy from the metal to the water.  This transfer will occur towards an equilibrium of thermal energy in the water and in the metal. Thus we can conclude that the amount of thermal energy lost by the metal will equal the amount of thermal energy gained by the water.  However we notice that the water increases by only 5^oC  and the metal decreases by 65^oC.  This is becasue of the difference of the specific heats of these substances.  The specific heat capacity of a substance is the heat required to increase the temperature of 1g of a substance by 1^oC. The metal can be conluded to have a smaller specific heat than the water because the same amount of energy transfer led to a much larger change in termperature for the metal as compared to the water.  

4.184 J/gK is the cited value for the specific heat of water and should be memorized. This is used during calorimeter calculations, specifically when using the equation q= mc delta(T).

This question involves the total energy needed for three different processes: the temperature raise from  to , the melting of the ice, and the temperature raise from  to . For the first and third transitions we will use the equation . For the melting of ice, we will use the equation .

1. 

2. 

3. 

Finally, we will need to sum the energy required for each step to find the total energy.