What step usually comes last when solving numeric problems

1 answer:

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.

You might be interested in

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 $PCl_3$