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Crank
3 years ago
7

An ideal gas cannot exist outside of ideal gas conditions true or false ?

Chemistry
1 answer:
zzz [600]3 years ago
5 0

Answer:

Yes

Explanation:

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Explain three path ways of respiration​
Phantasy [73]

Answer:

Explanation:

1:faringe:es un conducto que permite la comunicación entre las fosas nasales (el paso del aire por la nasofaringe a la laringe) y la cavidad bucal

Laringe:es una cavidad formada por cartílagos que presenta una saliente comúnmente llamada "nuez" en la laringe se encuentra las cuerdas vocales que al vibrar con el aire produce la voz

Tráquea:es un conducto de doce centímetros de longitud,situado delante del estómago.la tráquea brinda una vía abierta al aire que entra y sale se los pulmones

6 0
3 years ago
If the enthalpy value for a reaction is negative, what does that indicate about the reaction?
likoan [24]
This indicates that the reaction is exothermic meaning that it releases heat/energy
7 0
3 years ago
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How toxic is uranium
Serhud [2]

Answer:

Inhaling large concentrations of uranium can cause lung cancer from the exposure to alpha particles. Uranium is also a toxic chemical, meaning that ingestion of uranium can cause kidney damage from its chemical properties much sooner than its radioactive properties would cause cancers of the bone or liver.

Explanation:

6 0
3 years ago
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I NEED HELP ASAP!!<br> I DOMT KNOW HOW TO DO THIS
KatRina [158]
What part of it are you confused about
3 0
3 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
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