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pishuonlain [190]
3 years ago
8

`as much as we would like for it to be the case, we never get all i the product we are supposed to get when we run a real reacti

on. suppose you have a theoretical yield of a 50.0 grams of product. however, your actual yield is only 32.0 g. what is your percent yield?
Chemistry
1 answer:
zavuch27 [327]3 years ago
3 0
Need more information 
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Which answer provides the correct name for the following hydrocarbon?
ivanzaharov [21]
1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3

2) Only single bonds => alkane => sufix ane

3) no substitutions

4) 5 carbons = > prefix penta.

Therefore, the name is pentane.
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Marijuana most abundant psychoactive chemical is delta-9- tetrahydrocannabinol or
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<h2>Answer:</h2>

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Does anyone know the ones that are blank
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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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Potassium is a metal. Which property would you expect it to have?
bazaltina [42]

Answer: It’s a bad conductor of electricity

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