It says on google
<span>An acid-base indicator (e.g., phenolphthalein) changes color depending on the pH. Redox indicators are also frequently used. A drop of indicator solution is added to the titration at the start; when the color changes the endpoint has been reached, this is an approximation of the equivalence point.</span>
Answer:
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.
In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.
In this case:
- c= 4.184

- m=21 g
- ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C
Replacing:
Q= 4.184
* 21 g* (-6 C)
Q= - 527.184 J
To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.
<u><em>
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>
Answer is : It is a good thermal and electrical conductor. -The main point to be noted is that aluminium is a highly reactive element and still it is used for making cooking utensils. The reason is that aluminium has a very high affinity for oxygen. So, it reacts with oxygen and forms a layer of aluminium oxide on its surface.
Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
Answer:
about 79% (79.04369332 to be exact)
Explanation:
Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100
Br= 3 x 79.9/303.25 x100=79.04369332