Calculate the standard enthalpy change for the reaction →2Fe(s) + Al₂O3(s) 2Al(s) + Fe₂O3(s) given that 2Al(s) + O₂(g) 2Fe(s) + 30₂(g) Al₂O3(s) AH-1669.8 kJ/mol → Fe₂O3(s) AH = -822.2 kJ/mol rxn
1. Calculate the standard enthalpy change for the reaction →2Fe(s) + Al₂O3(s) 2Al(s) + Fe₂O3(s) given that 2Al(s) + O₂(g) 2Fe(s) + 30₂(g) Al₂O3(s) AH-1669.8 kJ/mol → Fe₂O3(s) AH = -822.2 kJ/mol rxn
Jawaban:
-3335.6 kJ/mol
Penjelasan:
Untuk menghitung perubahan entalpi standar dari suatu reaksi, kita dapat menggunakan persamaan ΔH = ΣΔHf(products) - ΣΔHf(reactants), dimana ΔH adalah perubahan entalpi standar, ΔHf adalah entalpi pembakaran standar, dan ΣΔHf(products) dan ΣΔHf(reactants) adalah jumlah entalpi pembakaran standar dari produk dan reaktan.
Dalam kasus ini, perubahan entalpi standar untuk reaksi →2Fe(s) + Al₂O3(s) 2Al(s) + Fe₂O3(s) adalah:
ΔH = ΣΔHf(products) - ΣΔHf(reactants)
= [2(-822.2 kJ/mol) + (-1669.8 kJ/mol)] - [0 kJ/mol + 2(-822.2 kJ/mol)]
= -3335.6 kJ/mol
2. tolong bantu kaawe use the bond......energy to calculate enthalpy of the reaction
of?
maaf kalo salah. ini materinya tentang apa ya kalo boleh tau? ada pilihannya ga?
3. Which equation represents the standard enthalpy change of formation of water .....
Jawaban:
equation represents the standard enthalpy change of formation of water
is 2H2(g)+O2(g)→2H2O(l)
4. mohon bantuannyawhen 8.0 g of metal (M) was oxidized in the oxygen atmosphere 10 g of MO was obtained owing the following reaction. Answer the following questions2M + O₂ ⇒ 2MO1. How many liter of O₂ were needed in this reaction? answer the volume at the standard state2. Calculate the weight percentage of M in MO3. Calculate the atomic weight of M
2M + O₂ ⇒ 2MO
8 g 2 g 10 g
0,125 mol 0,0625 mol
1) O₂ Volume = [tex] \frac{gram}{Mr}.22,4 = \frac{2}{32}.22,4= 1,4 [/tex] liter
3) mol = [tex] \frac{gram}{Ar} [/tex]
0,125 = [tex] \frac{8}{Ar} [/tex]
Ar M = [tex] \frac{8}{0,125} = 64 [/tex]
2) percentage M in MO
[tex] \frac{Ar M}{Mr MO} [/tex] x 100%
[tex] \frac{64}{80} [/tex] x 100% = 80%
5. Calculate the heat of reaction for combustion of C2H5OH using bond energies
Reaction C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O
Then draw the structural formulas for each molecule showing single and double bonds.
C₂H₅OH has one C-C bond, five C-H bonds, one C-O bond and one O-H bond
add up the bond energies for a mole of each of these single bonds from your bond energy chart.
Now in O₂ we have O=O double bond so add up the total bond energy for three moles of O₂. (three times the doubly bonded oxygen bond energy)
CO₂ is formed O=C=O, and you have two moles of CO₂, and four moles of C=O bonds so multiply the C=O bond energy by 4
and for H₂O, bonded H-O-H you will have 6 moles of H-O bonds, so multiply the H-O bond energy by six.
Now to get ΔH for this reaction, simply subtract the sum of the reactant bond energies from the sum of the product bond energies and you have it.
6. 5 ticket for the theatre cost 45$. Calculate the cost of 1,3,11
If the theather cost of 5 ticket is 45$
1 ticket 9$
3 ticket 27$
11 ticket 99$
7. Calculate SS, variance, and standard deviation for the following population of N 5 5 scores: 2, 13, 4, 10, 6. (Note: The definitional formula works well with these scores.)
aidil Fitri Acil Aldi ah
8. 28. Which does NOT change with time for a first-order reaction? (A) the amount of reactant that disappears in each half- life (B) the concentration of the reactant (C) the length of each half-life (D) the rate of the reaction
Pada reaksi order pertama yang tidak mengalami perubahan adalah :
C. The length of each half-life
Pembahasan
Reaksi orde pertama adalah reaksi dalam orde menetap, laju reaksi berbanding lurus dengan konsentrasi salah satu reaktan. Reaksi orde pertama sering memiliki bentuk umum A → produk. Tingkat diferensial untuk reaksi orde pertama adalah sebagai berikut:
laju=−Δ[A]/Δt=k[A]
Jika konsentrasi A digandakan, laju reaksi menjadi dua kali lipat, jika konsentrasi A dinaikkan dengan faktor 10, laju reaksi meningkat dengan faktor 10, dan seterusnya. Karena satuan laju reaksi selalu mol per liter per detik, satuan konstanta laju orde pertama adalah detik timbal balik (s−1). Hukum laju terintegrasi untuk reaksi orde pertama dapat ditulis dalam dua cara yang berbeda: satu menggunakan eksponen dan satu menggunakan logaritma. Ingat bahwa hukum laju terintegrasi memberikan hubungan antara konsentrasi reaktan dan waktu.
Pelajari lebih lanjut
Pelajari lebih lanjut mengenai materi laju reaksi orde pertama pada link berikut :
https://brainly.co.id/tugas/6386928
#BelajarBersamaBrainly
#SPJ1
9. why is the thermite reaction useful for welding rails?
Jawaban:
Thermite welding is widely used to weld railroad rails.The weld quality of chemically pure thermite is low due to the low heat penetration into the joining metals and the very low carbon and alloy content in the nearly pure molten iron.
10. what is the formula to calculate rate of reaction?
if you calculated for chemestry for reaction you must be know is formula of mole ,molaritas and valensi of ionic atom.
11. The rate constant for a first-order reaction is 3.64. 10 - 2 seconds -1 at 298 k.. what is the rate constant for the reaction at 350 k if the activation energy of the reaction is 50.2 kj/mol. given: the quantity r = 8.314 j/k.mol.
Jawaban terlampir di foto ya!
#GeniuzS05
#SAYIDI
12. If the mass is found as following = Copper : 0,52g H2O : 0,245g Chloride : 23.872g Calculate the empirical formula of substance in the reaction.
gangerti kak aku juga bingung
13. Calculate the distance travelled by the particle for the period of time
Jawab:
Penjelasan dengan langkah-langkah:
14. 3) 0.1 mol of lithium reacts with excess oxygen gas. (a) Write a balanced chemical equation for the reaction. (b) Calculate the mass of the product formed.
Jawaban:
Penjelasan:
(a) A balanced chemical equation for the reaction is:
2Li + O2 -> 2Li2O
(b) To calculate the mass of the product formed, we can use the molecular weight of lithium oxide (Li2O). The molecular weight of Li2O is 30.8 g/mol. Therefore, the mass of 0.1 mol Li2O is 3.08 grams.
15. Word equation for the reaction of astatine with iron
Jawaban:
Persamaan kata untuk reaksi astatin dengan besi
Penjelasan:
suruh apasih aku artiin aj
16. Write the word equation for the reaction between zinc and copper sulfate.
Jawaban:
Zn + CuSO4 ⇒ ZnSO4 + Cu
Penjelasan:
Zn has oxidation number equal to +2. Cu has oxidation number equal to +2. SO4 has oxidation number equal to -2.
17. 2. A reaction between 70.0 g of copper(II) oxide and 50 mL of 2.0 M nitric acid produces copper(II) nitrate, Cu(NO3)2 and water. (a) Write the balanced chemical equation for the above reaction. (b) Determine the limiting reactant. (c) Calculate the mass of excess reactant after the reaction. (66.02 g) (d) Determine the percentage yield if the actual mass of copper(II) nitrate obtained from the reaction is 8.5 g (90.62%)
Jawaban:
a. The balanced chemical equation for the reaction above
CuO + 2[tex]HNO_{3}[/tex] → [tex]Cu(NO_{3})_{2}[/tex] + [tex]H_{2}O[/tex]
b. he limiting reactant: nitric acid ([tex]HNO_{3}[/tex])
c. the mass of excess reactant after the reaction = 67,97 g
d. the percentage yield if the actual mass of copper(II) nitrate obtained from the reaction is 8.5 g = 50,47%
Penjelasan:
Persamaan reaksi dari reaksi tersebut adalah:
CuO + 2[tex]HNO_{3}[/tex] → [tex]Cu(NO_{3})_{2}[/tex] + [tex]H_{2}O[/tex]
mol CuO = [tex]\frac{massa}{Mr}[/tex]
= [tex]\frac{70}{79,5}[/tex]
= 0,88 mol
mol [tex]HNO_{3}[/tex] = V x M
= 50 mL x 2 M
= 100 mmol = 0,1 mol
Pereaksi pembatas dapat ditentukan dengan membagi mol reaktan dengan keofisienya. Reaktan yang hasil baginya lebih kecil merupakan pereaksi pembatas.
[tex]\frac{mol CuO}{koef CuO}[/tex] = [tex]\frac{0,88}{1}[/tex] = 0,88
[tex]\frac{mol HNO_{3} }{koef HNO_{3} }[/tex] = [tex]\frac{0,1}{2}[/tex] = 0,05
Hasil bagi yang lebih kecil adalah [tex]HNO_{3}[/tex] jadi pereaksi pembatasnya adalah [tex]HNO_{3}[/tex]. Sehingga:
mol CuO yang bereaksi = [tex]\frac{koef CuO}{Koef HNO_{3} }[/tex] x mol [tex]HNO_{3}[/tex]
= [tex]\frac{1}{2}[/tex] x 0,05
= 0,025 mol
mol [tex]Cu(NO_{3})_{2}[/tex] hasil reaksi = [tex]\frac{koef Cu(NO_{3})_{2} }{Koef HNO_{3} }[/tex] x mol [tex]HNO_{3}[/tex]
= [tex]\frac{1}{2}[/tex] x 0,05
= 0,025 mol
mol [tex]H_{2}O[/tex] hasil reaksi = [tex]\frac{koef H_{2}O }{Koef HNO_{3} }[/tex] x mol [tex]HNO_{3}[/tex]
= [tex]\frac{1}{2}[/tex] x 0,05
= 0,025 mol
CuO + 2[tex]HNO_{3}[/tex] → [tex]Cu(NO_{3})_{2}[/tex] + [tex]H_{2}O[/tex]
Awal : 0,88 mol 0,05 mol
Reaksi : 0,025 mol 0,05 mol 0,025 mol 0,025 mol
--------------------------------------------------------------------------------------------
Akhir : 0,855 mol - 0,025 mol 0,025 mol
Reaktan yang bersisa diakhir reaksi adalah CuO sebanyak 0,855 mol. Massa CuO = mol x Mr
= 0,855 x 79,5
= 67,97 g
[tex]Cu(NO_{3})_{2}[/tex] yang dihasilkan adalah 0,025 mol.
Massa [tex]Cu(NO_{3})_{2}[/tex] = mol x Mr
= 0,025 x 171,5
= 4,29 g
%[tex]Cu(NO_{3})_{2}[/tex] = [tex]\frac{4,29}{8,5}[/tex] x 100%
= 50,47%
Materi tentang pereaksi pembatas dapat disimak pada link:
https://brainly.co.id/tugas/40947892
#BelajarBersamaBrainly
18. calculate the total sales for the year ended 31 december 2010
hitung total penjualan untuk tahun berakhir 31 desember 2010
19. Write the complete balanced equation for the reaction between iron (III) oxide (Fe2O3) and water (H2O). You do not need to make the subscripts smaller; just write them out as regular numbers. For example: Fe2O3
Fe2O3 + 3 H2O ---> 2 Fe(OH)3
20. the reaction a Fe2O3+ b Fe2(SO4)3+ d H2o
and what am I supposed to do with this reaction? even it out?
0 komentar:
Posting Komentar