When an object/liquid/food etc contains hot energy, the heat particles are moving around causing the heat. 8ounces of soup would cool off faster, because there is less liquid for the heat particles to be in, there for less heat particles, therefore cooling off faster. Pls give Brainliest, and good luck!
Answer:
Explanation:
Hello,
In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:
The concentrations of the A and B ions in the solution are:
![[A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M](https://tex.z-dn.net/?f=%5BA%5D%3D0.099%20%5Cfrac%7BmolAB_3%7D%7BL%7D%2A%5Cfrac%7B1molA%7D%7B1molAB_3%7D%20%20%3D0.0099M)
![[B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M](https://tex.z-dn.net/?f=%5BB%5D%3D0.099%20%5Cfrac%7BmolAB_3%7D%7BL%7D%2A%5Cfrac%7B3molB%7D%7B1molAB_3%7D%20%20%3D0.000.297M)
Then, as the solubility product is defined as:
![Ksp=[A][B]^3](https://tex.z-dn.net/?f=Ksp%3D%5BA%5D%5BB%5D%5E3)
Due to the given dissociation, it turns out:
![Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}](https://tex.z-dn.net/?f=Ksp%3D%5B0.099M%5D%5B0.297M%5D%5E3%5C%5C%5C%5CKsp%3D2.59x10%5E%7B-3%7D)
Regards.
The specific heat capacity of unknown substance is 1.333 g/j/°c
<u><em> calculation</em></u>
Specific heat capacity is calculated using Q= MCΔT formula
where;
Q(heat)= 800 j
M(mass) = 30 g
C(specific heat capacity) =?
ΔT( change in temperature) = 40°c - 20°c = 20°c
make C the subject of the formula by diving both side of the formula by MΔT
C is therefore = Q/MΔT
800j / 30 g x 20°c = 1.333 j/g/°c
False. Water is basically better at sticking to itself than it is in sticking to the wax.
