Which of the following chemicals is a nonelectrolyte?a. molten naphthalenab. sodium hydroxide solutionc. aqueous sulfuric acid solutiond. aqueous ethanoic acid solutione. aqueous potassium iodide solution
1. Which of the following chemicals is a nonelectrolyte?a. molten naphthalenab. sodium hydroxide solutionc. aqueous sulfuric acid solutiond. aqueous ethanoic acid solutione. aqueous potassium iodide solution
The First thing that you must knew about Electrolyte Solution, Have the Criteria:
- The Solution that Perfect Ionized
- Strong Acid or alkali (have strong electrolyte) and weak Acid or alkali (have Weak Elektrolytes)
- Salt Solution (NaCl, KCl, and any others)
I thing Molten Napthalena Is the best Answer, Napthalena as an Organic coumpund has a Stabilized structur, so dificult to ionized.
I Hope that answer your question...
2. is an aqueous solution of sodium bicatbonate NaHCO2 acidic ,basic ,or neutral??
The anion undergoes hydrolysis to give free OH- ions and the acid in aqueous solution.
Reaction:
NaHCO₃ + H₂O → H₂CO₃ + OH⁻ + Na⁺
Since carbonic acid is a weak acid, it remains undissociated, thus the solution becomes basic due to the presence of additional OH- ions produced by hydrolysis.
SOLVED :)
3. an aqueous solution of sodium sulfate, naso, is electrolysed using carbon electrodes. what substance is formed at the cathode
Jawaban:
The substances formed at the cathode are hydrogen gas and hydroxide ions.
Tolong jadikan jawaban terbaik yaa, saling bantu
4. A 20 ml solution of 0,0512 M oxalic acid was used to standardise an unknown solution of potassium permanganate. An average of 21,24 ml of permanganate was required. Calculate the concentration of the potassium permanganate solution
Jawaban:
0.0964 M
Penjelasan:
H2C2O4 = KMnO4
V1 × N1 = V2 × N2
20 × 0.0512 × 2 = 21.24 × M × 1
[tex]M = \frac{20 × 0.0512 × 2 }{21.24} \\ M = 0.0964 [/tex]
5. 7. The pH of an aqueous solution of hydrochloric acid is 2.What is the pH of the solution after the addition of 10 g of sodium chloride?Nilai pH suatu larutan akueus asid hidroklorik ialah 2.Berapakah nilai pH larutan itu selepas penambahan 10 g natrium klorida?A 1B 2C 7D 9
Jawaban:
B 2
Penjelasan:
tetap 2 karna larutan yang dimasukin adlah larutan yang netral jadi ngk bakal ganggu apa apa
NaCl dari NaOH basa kuat HCl asam kuat jadi garam netral
6. 1,71 gram of non-electrolyte subtance is dissolved in 100 gram of water. The solution boils at a temperature of 100.026 C. If the elevantion constant of the water boiling point Kb is 0.52 kg/molal, teh molecular weight of that subtance is...g/mol
Tb - Tb° = (gr ÷ Mr) × (1000 ÷ p) × Kb
100,026 - 100 = (1,71 ÷ Mr) × (1000 ÷ 100) × 0,52
Mr = 200
7. the solution containing 6 g of CH3COOH and 4.1 g CH3COONa in 2 L of solution has a pH (Ar H = 1; C = 12; O = 16; Na = 23; Ka CH3COOH = 2x10^-5)
Jawaban:
pH = 5
Penjelasan:
m CH3COOH = 6 g
Mr CH3COOH = 60
n CH3COOH = 6/60 = 0,1 mol
m CH3COONa = 4,1 g
Mr CH3COONa = 82
n CH3COONa = 4,1/82 = 0,05 mol
V = 2 L
[H+] = Ka x (mol CH3COONa/mol CH3COOH)
= 2 x 10^-5 x (0,05/0,1)
= 2 x 10^-5 x 0,5
= 1 x 10^-5
pH = - log [H+]
= - log (1 x 10^-5)
= 5
8. compose a dialogue containing an expression of admiration..
a : You can rate how much repetition?
b : I can score 65, I have not been satisfied. How about you?
a : thank God I got a score of 100
b : wow congratulations, you're lucky you're so smart and deserve to be champion
a : amen. thanks a lot
9. comose a dialogue containing an expression of admiration
artinya Buat sebuah dialog yang berisi ekspresi kekagumanBuat sebuah dialog yang berisi ekspresi kekaguman
10. An indigestion (antacid) tablet containing 500 MG of Al(OH)3 is dissolved in 200 ml of water. if Kb of basic solution is 10pangkat - 5 and MrAl(OH) =78,the pH of the solution that is formed is.. A. 8.98 B. 9.98 C. 10.98 D. 11.98 E. 12.98
antasida -> al(oh)3
massa = 500 mg = 0,5 gr
v = 200ml = 0,2 lt
Mr = 78
Kb = 10^-5
Mb = gr/ Mr . v
= 0,5 / 0,2 x 78
=0,03205
[OH-] = √ kb . Mb
= √10^-5 x 0,03
= √0,3205 x 10^-6
= 0,56 x 10^-3
= 5,6 x 10^-4
= 56 x 10^-5
pOH = - log (OH-)
= - log (56 x 10^-5)
= 5 - log 56
pH = 14 - (5 - log 56)
= 9 + log 56
= 9 + 1,71
= 10,71
maaf gak ketemu... semoga membantu
11. What is the molarity of a solution containing 4 moles of KCl in 2.5 L of solution?
Penjelasan:
dik : n KCl = 4 mol
v = 2,5 L
dit : M = ?
jb :
M = n / v
= 4 mol / 2,5 L
= 1,6 mol/L
semoga membantu :)
12. the molality of solution containing 20% mass of ethanol (C6H5OH, Mr = 46 gram/mole) is.......molal... please jwb soal kimia aku :') materi uts
20% ethanol berarti pelarut nya 80%, kita ambil contoh itu larutan 100 gr berarti ethanol nya 20gr, tinggal di hitung pakai rumus n= m/Mr
n= 20/46= 0.434 mol
dik : m ( etanol) = 20 gr ( 100-20=80)
jadi , massa C₆H₅OH = 80
dit : m ...?
m = massa / Mr x 1000/P
= 20 / 46 X 1000 / 80
= 0,43 / 12,5
= 0.0344
13. water boils at 100°C and..........at 0°C
Jawaban:
Almost freeze
Penjelasan:
water boils at 100°C and almost freeze at 0°C
air mendidih saat suhu 100°C dan hampir beku saat suhu 0°C
14. 3,42 gram of non-electrolyte substance is dissolved in 200 gram of water. The solution boils at temperature of 100,026°C. If the boiling point elevation constan (Kb) of water is 0,52°C kg/mol the molecular weight of that sunstance is...
Mr = (gr terlarut /delta Tb) x (1000/gr pelarut) x Kb
Masukkan angka angkanya maka Mr = 342
15. The formulae of insoluble compounds can be found by precipitation reactions.To 12.0 cm of an aqueous solution of the nitrate of metal T was added 2.0 cm' of aqueous sodium phosphate,Na PO, The concentration of both solutions was 1.0 mol/dm'. When the precipitate had settled, its height wasmeasured.solutionprecipitate ofthe phosphateof metal Theight ofprecipitale#tolong bantu jawabb pleaseeeebener bener gapaham
Penjelasan:
wnwjwjenebebebsjsjwbbwbebebwbbwbwwbwbebebebbebebwbwbebwbwbwbwbwnwnwn maaf maaf sekali aku baru masuk brainlay tapi aku mau banyak poin
16. P.L.'s herbal sleeping potion was mixed into tea and taken at bedtime. The dissolved mixtureis called a(n) _______________________ and is taken at _______________________. A. Elixir and QAM b. Emulsion and bid c. Suspension and hs d. Aqueous solution and hs e. Aqueous solution and QAM
kenapa lama sekali aku mau main ff
17. water ... at 0 degree celsius and ...at 100 degree celsiusa,freezes-boilsb,boils-freezesc,freeze-boild, boils-freeze
Jawaban:
A. freezes-boils
Penjelasan:
Jawaban:
feezes and boils
Penjelasan:
water freezes at 0 degree Celsius and boils at 100 degree celsius
18. dialog gabungan containing an expresion of admiration
A: Are you, B?
B: Yes, i'm. Sorry, who are you?
A: Hey! I'm A! we went to high school together!
B: Uhm...... Are u, A? from the calculus class? Ah i remember you! You look so different now, A!
A: A lot of things change, B. Hey, i heard that you work for this top PR firm in Jakarta. Ah, i admire your job.
B: Don't say that kind of thing. I'm sure your job is pretty great too.
p.s: a sama b nya bisa diganti jadi nama orang kok
19. calculate the molality of 6.5 M aqueous solution of an acid , HA with a density of 0.888 g cm'3 .given the molar mass of acid is 98.0g/mol
Pembahasan:
Calculate the molality of 6.5 M aqueous solution of an acid , HA with a density of 0.888 g/cm³ .given the molar mass of acid is 98.0g/mol
Diketahui:
Molaritas HA = 6,5 M
ρ = 0,888 g/cm³
Mr = 98 g/mol
1 cm³ = 1 mL
Ditanya: molalitas...?
Jawab:
Pada soal ini kita diminta untuk mengkonversi molaritas suatu larutan asam menjadi molalitasnya. Pada dasarnya Molarita dan molaritas menyatakan hal yang sama yaitu konsentrasi zat dan suatu larutan, namun perbedaannya yaitu molaritas dinyatakan per satuan volume pelarut sedangkan molalitas per satuan massa pelarut yang digunakan.
Adapun rumus molaritas dan molalitas adalah sebagai berikut:
M = mol/Volume pelarut (L)
m = mol/Massa pelarut (Kg)
untuk itu kita harus mengetahui massa pelarut yang digunakan terlebih dahulu.
Berdasarkan soal diketahui asam HA memiliki molaritas 6,5 M namun tidak disebutkan volume larutanya, sehingga untuk menyelesaikan soal ini maka kita misalkan saja bahwa larutan asam HA 6,5 M tersedia sebanyak 1 L.
Sehingga bisa kita peroleh informasi sebagai berikut:
M = mol / V ==> mol = M . V
mol = 6,5 M . 1 L = 6,5 mol
Jadi mol asam HA 6,5 M sebanyak 1 L adalah 6,5 mol
Selanjutnya untuk menghitung massa pelarut yang digunakan kita bisa mencari massa larutan berdasarkan nilai densitas yang diberikan, yaitu:
Massa Larutan = Volume larutan x ρ
Massa Larutan = 1 L x 1000 x 0,888 g/cm³ ==> 1 L = 1000 mL
Massa Larutan = 888 gram
Selanjutnya kita hitung massa zat terlarut nya yaitu menggunakan data molnya,
mol = massa/ Mr ==> Massa = mol x Mr
Massa = 6,5 mol . 98 g/mol
Massa = 637 gram
Selanjutnya dari hasil perhitungan di atas kita bisa mengetahui massa pelarut, yaitu:
Massa Larutan = Massa Zat Terlarut + Massa Pelarut
Massa Pelarut = Massa Larutan - Massa Zat Terlarut
Massa Pelarut = 888 gram - 637 gram
Massa Pelarut = 251 gram = 0,251 Kg
Sehingga molalitas asam HA bisa kita cari sebagai berikut:
Molalitas HA = mol / Massa Pelarut
Molalitas HA = 6,5 mol / 0,251 Kg
Molalitas HA = 25,89 molal
Jadi molalitas dari asam HA 6,5 M dengan densitas 0,888 gram/cm³ adalah sebesar 25,89 molal.
Pelajari soal-soal Stoikiometri lainnya melalui link-link berikut:
https://brainly.co.id/tugas/16116985
https://brainly.co.id/tugas/16109125
https://brainly.co.id/tugas/16076344
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Selamat Belajar dan Tetap Semangat!!!
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Kelas : X
Mapel : KIMIA
BAB : Stoikiometri
Kata Kunci : Molaritas, molalitas, densitas, massa, larutan
Kode : 10.7.7.
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20. Water boils at what degree?
Water boils at 100 celcius
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