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lilavasa [31]
3 years ago
7

Please Help me if you can!:) i appreciate anything.

Chemistry
1 answer:
enot [183]3 years ago
7 0

Explanation:

1

Number of nucleon =

Molarmassofnucleon

Massofatom

=

1.6726×10

−24

g/nucleon

3.32×10

−23

g

=19.8=20(approximately)

It is given that element comprises of 2 atoms

Hence,number of nucleon = 2×20=40

2

You have 4.70 mol H2O

There are two H atoms in 1 molecule H2O.

Therefore, there must be 2*4.70 = 9.40 mols H in 4.70 mols H2O.

How many mols O in 4.70 mols H2O? That's 4.70 mols, of course.

Said another way, you have 2 mols H for every 1 mol H2O and 1 mol O for every 1 mol H2O.

So for 50 mols H2O you have 100 mols H and 50 mol O.

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Complete and balance the molecular equation for the reaction of aqueous iron(III) nitrate, Fe ( NO 3 ) 3 , and aqueous lithium h
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Answer:

The balanced chemical reaction is given as:

Fe(NO_3)_3(aq)+3LiOH(aq)\rightarrow Fe(OH)_3(s)+3LiNO_3(aq)

Explanation:

When aqueous ferric nitrate is allowed to react with aqueous lithium hydroxide it gives solid precipitate of ferric hydroxide and aqueous solution of lithium nitrate. The balanced chemical reaction is given as:

Fe(NO_3)_3(aq)+3LiOH(aq)\rightarrow Fe(OH)_3(s)+3LiNO_3(aq)

According to reaction , 1 mole of ferric nitrate reacts 3 moles of lithium hydroxide to give 1 mole of ferric hydroxide and 3 moles of lithium nitrate.

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I have a lab report I have to do for Chemistry on Edge, the lab is on Enthalpy.
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Answer:

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Explanation:

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Compare the boiling points of 1-pentyne and 1-octyne. Compare the vapor pressure curves of 1-butene and 1-heptene.
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Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn=
IgorC [24]

Answer: \Delta H^0=+0.3kJ/mol.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

S_{rhombic}+O_2(g)\rightarrow SO_2(g)    \Delta H^0_1=-296.06kJ   (1)

S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ  (2)

The final reaction is:  

S_{rhombic}\rightarrow S_{monoclinic}  \Delta H^0_3=?   (3)

By subtracting (1) and (2)

\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ

Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ

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Helppp science *10 points*
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4 th one is Midwest direction. Found from visible paragraph. Pls apload the content to answer
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