3. If a sound is louder or a light is brighter, what does this mean about the amplitude of

1 answer:

7 0

Answer: for 3. I would say it means that the amplitude is greater because if a sound is louder that means the amplitude is larger, same with the light.So that would mean the softer the sound the smaller the amplitude same with the dim of the light so for 4. It would be the amplitude is smaller. Hope this helps! :) ( if you have any questions feel free to ask)

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Hazardous materials are grouped into classes identifying their.

saveliy_v [14]

Hazardous materials are grouped into classes identifying their similarities in composition and structure.

<h3>Why hazardous materials are grouped into classes?</h3>

The hazardous materials are grouped into classes in order to tell us about the severity of hazard and it is done on the basis of similarity in composition.

So we can conclude that hazardous materials are grouped into classes identifying their similarities in composition and structure.

Learn more about hazardous here: brainly.com/question/7310653

8 0

1 year ago

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:

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