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>
It’s oxidation number is +6
Answer:
63.36gallons
Explanation:
Given:
Volume of water used for dialysis = 2.4 x 10²L
Solution:
We are to convert from liters to gallons.
The conversion factor is shown below:
1L = 0.264gallons
To convert to litre:
since 1L = 0.264gallons
2.4 x 10²L = (2.4 x 10² x 0.264)gallons; 63.36gallons
1. Take 100ml of water as solvent and boil it few minutes.
2. Now add one tea spoon sugar, one tea spoon tea leaves and 50ml of milk. Here sugar, tea leaves and milk are solute.
3. Now boil it again for few minutes so that sugar will dissolves in solution as sugar is soluble in water
4. Now filter the solution. Collect the filtrate in cup. The insoluble tea leaves will be left behind as residue.