TRUE or FALSE. adding a solute like sodium chloride to water will always increase the freezing point
2 answers:
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Answer:
Explanation:
Hello,
In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:
Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:
Clearly, it turns out negative since the concentration is diminishing due to its consumption.
Regards.
Taking into account the definition of calorimetry, the specific heat of metal is 0.165 .
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
- Q is the heat exchanged by a body of mass m.
- C is the specific heat substance.
- ΔT is the temperature variation.
In this case, you know:
For metal:
- Mass of metal = 50 g
- Initial temperature of metal= 45 °C
- Final temperature of metal= 11.08 ºC
- Specific heat of metal= ?
For water:
- Mass of water = 250 g
- Initial temperature of water= 10 ºC
- Final temperature of water= 11.08 ºC
- Specific heat of water = 1.035
Replacing in the expression to calculate heat exchanges:
For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)
For water: Qwater= 1.035 × 250 g× (11.08 C - 10 C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:
- Qmetal = + Qwater
- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 × 250 g× (11.08 C - 10 C)
Solving:
- Specific heat of metal× 50 g× (-33.92 C)= 1.035 × 250 g× 1.08 C
Specific heat of metal× 1696 g×C= 279.45 cal
Specific heat of metal=
<u><em>Specific heat of metal= 0.165 </em></u>
Finally, the specific heat of metal is 0.165 .
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