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3 years ago

How do electrons in the same orbital move?

Chemistry

1 answer:

Sladkaya [172]3 years ago

5 0

Two electrons !!!!!!!

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Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

Answer: The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

For process 2:

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

For process 3:

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

8 0

3 years ago

The information below describes a redox reaction.

lord [1]

Answer: option 1) 2Cr3+(aq)+6Cl-(aq) ------> 2Cr(s)+3Cl2(g)

Explanation:

1) Write the oxidation half-reaction:

$2Cl^-(aq)---\ \textgreater \ Cl_2(g)+2e^-$

2) Write the reduction half-raction:

$Cr^{3+}(aq)+3e^{-}---\ \textgreater \ Cr(s)$

3) Multiply each half-reaction by the appropiate coefficient to equal the number of electrons of both half-reactions.

$6Cl^{-}(aq)---\ \textgreater \ 3Cl_2(g)+6e^{-}

2Cr^{3+}(aq)+6e^{-}---\ \textgreater \ 2Cr(s)$

4) Add both half-reactions

$2Cr^{3+}+6Cl^{-}(aq)---\ \textgreater \ 2Cr(s) +3Cl_2(g)$

And that is the answer. You can count the atoms and charges on every side and check they are equal.

5 0

3 years ago

What is called what a scientist personal opinion affects the way experimental results are reported?

e-lub [12.9K]

The answer would be A.Bias because the scientist can form a Bias opinion based on his beliefs

3 0

3 years ago

Read 2 more answers

What is the Net Ionic equation for this chemical reaction: FeBr2+Na2S=FeS+2NaBr

Artyom0805 [142]

Answer: Fe(aq)+S(aq)=FeS(s)

Explanation: The Sodium and Bromine are spectator ions because they don't react with anything, you can see this by writing the ionic equation like so:

1.) Molecular formula (given): FeBr2 (aq)+Na2S (aq)= FeS(s)+2NaBr(aq)

Each dissolved FeBr2 breaks up into one Fe with a charge of 2+ and two Br with a negative charge. This gives you:

Fe(aq)+ 2Br(aq)+Na2S(aq)=FeS(s)+2NaBr

2.) Now repeat what was shown with the other compounds in the given molecular formula, and pay attention to the states that each ion is in (solid, liquid, aqueous, gas) because this will give you the ionic equation, which from there you can get rid of any ions that don't change amount or state.

3.) Ionic formula: Fe(aq)+ 2Br(aq)+2 Na(aq)+S (aq)=FeS(s)+2 Na(aq)+2Br(aq)

4.)When you've derived a total ionic equation (above), you'll find that some ions appear on both sides of the equation in equal numbers. For example, in this case two Na cations and two Br anions appear on both sides of the total ionic equation. What does this mean? It means these ions don't participate in the chemical reaction. They're present before and after the reaction. Nothing happens to them. So those are removed and you're left with the net ionic: Fe(aq)+S(aq)=FeS(s)

Hope this helps :)

7 0

3 years ago

How did the different reform movements we've discussed change and shape America? Pls help

tigry1 [53]

Answer:

ok ............bu am not so sure

6 0

3 years ago