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zavuch27 [327]
2 years ago
14

The compound adrenaline contains 56.79% c, 6.56% h, 28.37% o, and 8.28% n by mass. what is the empirical formula for adrenaline?

chempadhelp
Chemistry
1 answer:
Phoenix [80]2 years ago
8 0
To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:<span>
         mass        moles
C       56.79        4.73
H       6.56          6.50
O       28.37        1.77
N      8.28           0.59

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

</span>         moles                 ratio
C       4.73     / 0.59       8
H       6.50     / 0.59      11
O       1.77     / 0.59       3
N        0.59    / 0.59       1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
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3 0
3 years ago
The average rate of consumption of br− is 1.86×10−4 m/s over the first two minutes. what is the average rate of formation of br2
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By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation 
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
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7 0
2 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

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

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

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