Answer:
Honey is an homogeneous mixture
Salsa is an heterogeneous mixture
Explanation:
Honey is a sweet uniformly colored liquid that can be of a dark variety or of clear golden color. Honey is made in nature by bees from flower nectar and is used as a food additive or sweetener
Honey is a homogeneous mixture because the concentration of the components of honey are uniformly distributed throughout the mixture. Every portion has the same concentration of components
Salsa is a sauce made by mixing chopped tomatoes, onions, chilies, lime juice and seasoning and therefore consists of both solid and liquid components mixed in varying proportion such that part will have continuous that comes in between different types of solid and no two parts have exactly the same composition
Therefore, salsa is an heterogenous liquid.
After some thinking I have come to the conclusion that the answer is C.
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
<u>Answer:</u> The enthalpy of the reaction for the production of 
is coming out to be -74.9 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_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%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_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of is coming out to be -74.9 kJ
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
