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andriy [413]
3 years ago
15

What step usually comes last when solving numeric problems

Chemistry
1 answer:
Lilit [14]3 years ago
6 0
No mathematics the most common way of solving an equation will be PEMDAS so the answer for your question will most likely be subtraction.
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Why do you think that the electrons are not shown?
snow_lady [41]
Because there to tiny to see
7 0
3 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

6 0
3 years ago
PLZ HELP IM GIVING 50 FRICKING POINTS. NO WRONG ANSWERS PLZ
Liula [17]

Answer:

I think it's B

Explanation:

I dont have much experience with the periodic table, but I just think its B.

3 0
3 years ago
If 5 moles of P4 reacted with 22 moles Cl2 according to the above reaction, determine:
GaryK [48]

Solution:

P_4+6Cl_2\rightarrow 4PCl_3

a) By stoichoimetry If 6 moles of Cl_2 gives 4 mole of PCl_3 , then 22 moles of Cl_2 will give: \frac{4}{6}\times 22=\frac{44}{3}=14.66 moles of PCl_3

b) If 6 moles of Cl_2 reacts with one mole of P_4 , then 22 moles of Cl_2 will react : \frac{1}{6}\times 22=\frac{11}{3}=3.66 moles of P_4

Moles of P_4  left after the reaction = 5-3.66=\frac{4}{3}=1.34 moles.

c) 0 Moles of Cl_2 , since it is present in less amount it will get completely consumed in the reaction. Hence, it is limiting reagent.



7 0
3 years ago
Material use make car tyres and chewing gum?
Ulleksa [173]

Answer:

Rubber

Rubber

Make me brainlyist

3 0
3 years ago
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