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Veseljchak [2.6K]
2 years ago
14

How is a mixture such as a plate of pasta different from a mixture such as air

Chemistry
1 answer:
stealth61 [152]2 years ago
6 0

Answer:

Well there is a lot of differences between the two. Its called homogeneous and Heterogeneous mixtures. Homogeneous mixtures are all the substances are evenly distributed throughout the mixture (salt water, air, blood).  Heterogeneous mixtures are the substances that are not evenly distributed (chocolate chip cookies, pizza, rocks). So Pasta with sauce and meatballs is heterogeneous and air is homogeneous




HOPE THIS HELPS HAVE A GREAT DAY!!~

Explanation:

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The pure substance of water mixes with the waste. The waste does not dissolve into the water, therefore it is a mixture.

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Consider the reaction between ozone and a metal cation, m2+, to form the metal oxide, mo2, and dioxygen: o3 + m2+(aq) + h2o(l) →
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Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)
vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

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 \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

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Atomic number is equivalent to its protons and electrons. :)
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