The large piece of jewelry that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml
To solve this problem the formulas and the procedures that we have to use are:
Where:
- d= density
- m= mass
- v= volume
- v(f) = final volume
- v(i) = initial volume
Information about the problem:
- m = 132.6 g
- v(i) = 48.6 ml
- v(f) = 61.2 ml
- v = ?
- d =?
Applying the volume formula we get:
v = v(f)-v(i)
v = 61.2 ml - 48.6 ml
v = 12.6 ml
Applying the density formula we get:
d = m/v
d = 132.6 g/12.6 ml
d = 10.523 g/ml
<h3>What is density?</h3>
It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.
Learn more about density in: brainly.com/question/1354972
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<span>The characteristics of a wave are given certain names to describe them, and this helps scientists to accurately describe any given wave. The characteristic known as frequency describes the number of waves that pass a point, and it is measured in waves per second, or Hertz, which is given the symbol Hz, but can also be described using the inverse of the SI unit for second, s^-1.</span>
<u>Answer:</u> For the given equation, only iron has the value of
equal to 0 kJ.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%5D-%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe_2O_3%28s%29%29%7D%29%5D)
The enthalpy of formation for the substances present in their elemental state is taken as 0.
Here, iron is present in its elemental state which is solid.
Hence, for the given equation, only iron has the value of
equal to 0 kJ.