Wawasan Genius

the mole fraction of solute and solvent if 15 gram of accetic acid (CH3COOH)is dissolved in 90 gram of water is

1. the mole fraction of solute and solvent if 15 gram of accetic acid (CH3COOH)is dissolved in 90 gram of water is

nt = massa terlarut/mr terlarut
nt = 15/60
nt = 0,25 mol

np = massa pelarut/mr pelarut
np = 90/18
np = 5 mol

fraksi mol pelarut
xp = np/nt+np
xp = 5/5+0,25
xp = 0,9524

fraksi mol terlarut
xt = 1-xp
xt = 1-0,9524
xt = 0,0476solute 0.05 dan solvent 0.95

2. determine the mole fraction of solute and solvent if 15 gram of acetic acid (CH3COOH) is dissolved in 90 gram of water?

n CH₃COOH = gr/Mr = 15/60 = 0,25 mol
n H₂O = gr/Mr = 90/18 = 5 mol
n solution = 0,25 + 5 = 5,25 mol

mole fraction of solute
[tex]Xt= \frac{n~CH_3COOH}{n~solution} \\ \\ = \frac{0,25}{5,25}=0,048 [/tex]

mole fraction of solvent
1-0,048 = 0,952

3. Determine the mole fraction of solute and solvent if 15 gra of acetic acid (CH3COOH) is dissolved in 90 gram of water

n acetic acid= m/Mr= 15/60=0,25
n water = m/Mr= 90/18=5
Solute fraction = 0,25/(0,25+5)= 0,25/5,25
Solvent fraction = 5/(0,25+5) = 5/5,25

4. 3.0 g of magnesium was added to 12.0g of ethanoic acid. Mg + 2CH 3 COOH → (CH3COO)2Mg + H 2 The mass of one mole of Mg is 24g. The mass of one mole of CH3COOH is 60g. Which one, magnesium or ethanoic acid, is in excess? You must show you nasoning ​

mass of Mg = 3 g

Atomic relative Mg (Ar) = 24 g/mole

mass of CH3COOH = 12 g

Molecular relative of CH3COOH (Mr) = 60 g/mole

Reaction occurs:

Mg + 2CH3COOH → (CH3COO)2Mg + H2 (balance)

first, calculate mole for each reactant

[tex]n \: = \frac{mass}{Ar/Mr} [/tex]

[tex]n \: Mg= \frac{3 \: g}{24 \: g/mole} = 0.125 \: mole [/tex]

[tex]n \: CH _{3}COOH = \frac{12 \: g}{60 \: g/mole} = 0.2 \: mole [/tex]

identifying the limiting reactant from balanced equation, use the mole from calculation before and use coefficient for each reactant from balanced equation

[tex]limiting \: reactant = \frac{mole \: reactant}{coefficient \: of \: reactant} [/tex]

[tex]limiting \: Mg = \frac{0.125}{1} = 0.125 \: mole[/tex]

[tex]limiting \: CH_{3}COOH = \frac{0.2}{2} = 0.1 \: mole[/tex]

use CH3COOH as limiting reactant because it has smallest mole quantity, so for 1 coefficient of each compoundcompound is 0,1 mole.

calculate the remaining reactant

Mg + 2CH3COOH → (CH3COO)2Mg + H2

initial 0.125 0.2 - -

reacted 0.1 0.2 0.1 0.1

excess 0.025 - 0.1 0.1

calculate mass of excess reactant, Mg

[tex]mass \: = excess \: mole \: \times Ar = 0.025 \: mole \times 24 \: g/mole = 0.6 \: g[/tex]

5. Di antara larutan-larutan berikut yang isotonik dengan larutan bacl2 0 1 m adalah

Jawaban:

NH CL

Penjelasan:

MAAF KALO SALAH SOALNYA SAYA JUGA KURANG PAHAM

6. Apa perbedaan elektrolisis larutan BaCl2 dengan elektrolisis leburan BaCl2?

Larutan : pada katoda yg tereduksi adalah air.

7. 40. Bacalah petikan teks fabel berikut!Kiki Kelinci, Bee Belalang, serta Yaya Ayam, dan beberapa binatang lain merasa malu kepadaMole. (2) Selama ini mereka sering meremehkan Mole. (3) Namun, hari ini Mole telahmenyelamatkan mereka dari pemburu. (4) Untung saja Mole adalah tikus yang baik hati. (5) Diatak pernah dendam kepada siapa pun yang telah menyakitinya.Karakter Mole yang dapat kita teladani terdapat pada kalimat nomor....a. (1) (2) (3)C. (3) (4) (5)b. (2) (3) (4)d. (1) (4) (5)​

Jawaban:

c. (3) (4) (5)

Penjelasan:

(3) mole menyelamatkan mereka dari pemburu

(4) mole adalah tikus yg baik hati

(5) dia tak pernah dendam kepada siapapun

8. Is-the-dug-the ground-mole-out of-usually

The mole is usually dug out of the ground.

___________
#Semoga bermanfaat ^_^Answer:

The mole is usually dug out of the ground

Thanks

9. Berapa Energi Kinetik Translasi dari satu mole suatu Gas Ideal pada suhu 0°C

Jawaban:

Dalam artikel ini akan membahas tentang teori kinetik gas secara keseluruhan, mulai dari pengertian gas ideal, persamamaan umum gas ideal, persamaan keadaan gas ideal, tekanan gas ideal, energi kinetik gas ideal, dan energi dalam gas ideal. Yuk, simak pembahasan lengkap tentang teori kinetik gas di bawah ini!

10. the mole fraction of solute and solvent if 15 gram of acetic acid (ch3cooh) is dissolved in 90 gram of water is

i dunno how to speak chemistry with English but I'll try to answer in Bahasa.

Diketahui :
nt =15 gram
np= 90 gram
Ditanya : Xp and Xt?
Penyelesaian :
Xp = np/np+nt = 90/90+15 = 0,85714286

Xp + Xt = 1
Xt = 1 - 0,85714286 = 0,14285714

11. Which one of the following molten electrolytes requires only one mole of electrons to produce one mole of metal atoms during electrolysis? A.calcium glories B. Iron (111)glories C. Magnesium bromide D.potassium bromide

jawabannya adalah c magnesium bromide

12. The heat of combustion of ethane gas is 373 kcal/mole. Assuming that б0.0 percent of the heat is useful, how many liters of ethane, measured at standard temperature and pressure, must be burned to convert 50.0 kg of water at 10.0 ° C to steam at 100.0 °C? One mole of a gas occupies 22.4 liters at precisely 0 °C and 1 atm.​

Jawaban:

3143.28 liters  of ethane gas must be burned .

Penjelasan:

m = 50 kg

η = 60%

c = 4200 J/kg°C

L = 2.260.000 J/kg

Kalor Untuk memanaskan air dari 10  °C menjadi 100 °C

Q₁ = m.c.ΔT

Kalor Untuk menguapkan air dari 100  °C menjadi uap 100 °C

Q₂ = m. L

Kalor Untuk memanaskan air dari 10  °C menjadi uap 100 °C

Q = Q₁ + Q₂

Q = m.c.ΔT + m. L

Q = 50. 4200. (100 - 10) + 50. 2.260.000

Q = 131900000 J

Q ini merupakan 60% dari kalor pembakaran gas etana awal.

Kalor pembakaran gas etana awal

Qa = 100/60 . 131900000

Qa = 219833333,3 J

Panas pembakaran gas etana 373 kcal/mole

= 373000 cal/mole

=373000 . 4,2 J/mole

= 1566600 J/mole

Jumlah mol gas etana yg dibakar n

n = 219833333,3/1566600 = 140,325 mol

Jumlah liter gas etana = n . 22,4 l

= 140,325 . 22,4

= 3143.28 liters

13. 100.0-g sample of a brand of rubbing alcohol contrains 65,0g of isopropyl alcohol C3 H7 OH and 35,0g of water. Whats is the mole fraction of isopropyl alcohol in the solution? What is the mole fraction of water?

Semoga dapat membantu ^^

14. 0.Tugom bigunnahSurat Al Falaq terdiri dari .... ayat. Turun dikota....5 ayat dan Mole​

Jawaban:

5 ayat

turun dikota mekkah

maaf kalo salah

semoga membantu...

Jawaban:

al falaq makiyyah berati turun di kota makkah terdiri dr 5 ayat termasuk salah satu 3 qul yg dalam cerita digunakan untuk witir rasulullah saw

Penjelasan:

semoga bermanfaat

15. . In RP A of Dentistry, Airlangga University there were pieces of Arsenate element. A student once tried to react between Oxygen and Arsenate in acidic condition and he observed what substance would be produced. In the end of the experiment, he found molecules of Hydrogen Arsenate (HAsO2) and water. O2 + As ---> HAsO2 + H2O. Determine how many As he used! (A)3 mole (B) 2 mole (C) 8 mole (D)4 mole (E) 6 mole

If want to know how many As is used in the reaction. We have to balance the reaction between O2 and As. Based on the reaction all of the elements lead to oxidation and reduction. We have to find out of each elements whether either oxidation or reduction.

Based on the reaction the O2 leads to reduction. So, we can write an equation reaction of the O2.

O2 → H2O

The As leads to reduction. So, we can write the reaction of As.

As → HAsO2

By using half-reaction method. We can balance redox reaction.

Red : O2 + 4H+ + 4e → 2H2O | x 3
Oks : As + 2H2O → HAsO2 + 3H+ 3e | 4

To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

4As + 3O2 + 8H2O + 12H+ + 12e- → 4HAsO2 + 6H2O + 12H+ + 12e-

Simplify the equation. The same species on opposite sides of the arrow can be canceled.
So, we can write the balance of whole coefficients reaction

4As + 3O2 + 2H2O → 4HAsO2

Based on the reaction. We know that there are 4 coefficient of As, So, in the reaction between O2 and As. It needs 4 mole of As to completely the reaction.

Wish it hepled
Keep Learning

Footnote :
periodni.com/half-reaction_method/

@XeMore

16. The mole fraction of solute and solvent if 15 gram of acetic acid (CH3COOH) is dissolved in 90 gram of water is

solute: acetic acid (CH₃COOH, Mr = 60)
solvent: water (H₂O, Mr = 18)

mole of solute: mass / Mr = 15 gr / 60 = 0,25
mole of solvent: mass / Mr = 90 gr / 18 = 5

X solute = moles of solute / total moles of solution
X solute = 0,25 / (0,25 + 5)
X solute = 0,0476

X solvent = 1 - X solue
X solvent = 1 - 0,0476
X solvent = 0,9524

17. Apa perbedaan elektrolisis larutan BaCl2 dengan elektrolisis leburan BaCl2?

Kalau leburan itu tidak ada zat pelarut di dalamnya. Contohnya ketika BaCl2 (Barium Klorida) yang merupakan garam ketika dipanaskan maka garam akan menguap dan mencair, hal inilah yang disebut leburan. Yang di elektrolisis adalah kation dan anion nya (apabila menggunakan logam inert)

Sementara larutan BaCl2 yaitu dengan cara dilarutkan (dalam pelarutan terdapat zat pelarut dan terlarut). Contohnya ketika kita melarutkan BaCl2 dengan air, hal ini disebut larutan BaCl2. Kalau dalam bentuk larutan, kation akan direduksi apabila berasal dari golongan I A, II A, III A, Al dan Mn. Jika bukan berasal dari golongan itu, maka yang direduksi adalah air.

Anion akan dioksidasi apabila suatu zat tidak berasal dari sisa asam oksigen. Jika anion berasal dari asam oksigen maka air yang akan mengalami oksidasi

18. bagaimana menulis not balok kres and mole 1 sampai kres and mole 7?

1BSR,2BSR-7BSR
.  .      .
1 2-    10

19. sebutkan 5 contoh moleku non logam

hidrogen, oksigen, karbon, nitrogen, heliumNama Indonesia = Nama Latin
belerang  = sulfur
bromin =bromium
fluorin = fluorine
fosforus =phosphorus
helium = helium

20. gambarkan bentuk moleku dari:a). sf4b). BrF 5

semoga membantu .....