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. Color reaction of ascorbic acid with sodium hydroxide stability ?
Jawaban:
Tanya ama gurunya saya bukan suhu
YNTKTS
Jawaban:
maaf yah..................
3. How many atoms of aluminum hydroxide may be formed from 65 grams of sodium hydroxide?
Jawaban:
tidak tau tidak tau tidak tau
4. Lithium reacts with water to form lithium hydroxide. What mass of lithium is required to produce 12 g of lithium hydroxide? (A) 2.0 g (B) 3.5 g (C) 7.0 g (D) 12 g
Jawaban:
b. 3,5 gram
Penjelasan:
dengan nilai koefisien reaksi kita tahu perbandingan mol reaktan dan produk
jawaban terlampir, semangat belajarnya
5. A student was given a sample of hydrochloric acid and asked to find its concentration in mol / dm. She titrated 25 cm' of the acid against a standard solution of sodium hydroxide of 0.1 mol dm. The average titre of sodium hydroxide was 30 cm. What is the concentration of the acid? A. 0.50 mol dm B. 0.12 mol dm C. 1.25 mol dm D. 1.0 mol dm
Jawaban:
c. 1.25 mol
Penjelasan:
semoga membantu
6. If sulphuric acid and sodium hydroxide solution are mixed, what is the result? Write your answer in the form of word equation
NaOH+H2SO4=>Na2SO4+H2O
setarakan
2NaOH+H2SO4=>Na2SO4+2H2O
7. 0.08 moles of sulfuric acid in 2 litres of solution has a Ph?
n (moles) = M / V
M (concentracion) = n x V = 0,08 moles x 2 Litres = 0,16 M
Since H₂SO₄ is strong acid, we must use this formula:
[H⁺] = acid valence x Concentration
[H⁺] = 2 x 0,16 M
[H⁺]= 0,32 M = 32 x 10⁻² M
pH = - log [H⁺]
pH = - log (32 x 10⁻² M)
pH = 2 - log 32
~apply some logarithm properties~
pH = 2 - 5 log 2
pH = 0,5
That gives us pH of said acid is 2-5log2.
I hope this helps~
8. 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
#backtoschoolcampaign
Selamat Belajar dan Tetap Semangat!!!
------------------------------------------------------------------------------------------------------
Kelas : X
Mapel : KIMIA
BAB : Stoikiometri
Kata Kunci : Molaritas, molalitas, densitas, massa, larutan
Kode : 10.7.7.
------------------------------------------------------------------------------------------------------
9. find the volume of sodium hydroxide 0.1 M to neutralize 20 ml of sulfate acid 0.2M
Jawab :
Volume natrium hidroksida (NaOH) 0,1 M yang dibutuhkan untuk menetralkan 20 mL asam sulfat (H₂SO₄) 0,2 M adalah 40 mL.
Penyelesaian Soal :
Diketahui : [NaOH] = 0,1 M
[H₂SO₄] = 0,2 M
V H₂SO₄ = 20 mL
Ditanya : V NaOH ?
Jawab :
V₁ × M₁ = V₂ × M₂
20 mL × 0,2 M = V₂ × 0,1 M
V₂ = (20 mL × 0,2 M) / 0,1 M
V₂ = 4/ 0,1 mL
V₂ = 40 mL
∴ Kesimpulan banyaknya volume NaOH adalah 40 mL.
Pembahasan :
Pengertian Molaritas (M)
Konsentrasi merupakan jumlah zat terlarut yang ada terhadap jumlah pelarut tertentu atau terhadap jumlah larutan tertentu. (diasumsikan zat terlarut berwujud cair atau padat, sedangkan pelarutnya berwujud cair) Konsentrasi dapat diungkapkan dengan beragam cara, salah satunya yang paling sering dipakai, dan memang akan kita gunakan sekarang ini adalah Molaritas (M), atau konsentrasi molar. Molaritas adalah jumlah mol terlarut setiap liter larutan. Atau bias diungkapkan dengan rumus:
M = n/V
Keterangan : M = konsentrasi zat
n = mol zat terlarut
V = volume larutan (L)
Pengenceran
Proses pengenceran adalah proses yang dilakukan dengan mencampurkan pelarut yang lebih ukurannya dari suatu zat yang dilarutkannya. engenceran pada prinsipnya hanya menambahkan pelarut saja, sehingga jumlah mol zat terlarut sebelum pengenceran sama dengan jumlah mol zat terlarut sesudah pengenceran. Dengan kata lain jumlah mmol zat terlarut sebelum pengenceran sama dengan jumlah mmol zat terlarut sesudah penegenceran atau jumlah massa zat terlarut sebelum pengenceran sama dengan jumlah massa zat terlarut sesudah pengenceran.
Rumus :
V₁ × M₁ = V₂ × M₂
Keterangan :
M₁ = Molaritas larutan sebelum pelarutan/pengenceran
V₁ = Volume larutan sebelum pelarutan/pengenceran
M₂ = Molaritas larutan sesudah pelarutan/pengenceran
V₂ = Volume Molaritas larutan sesudah pelarutan/pengenceran
Pelajari lebih lanjut :
materi tentang pH asam Basa https://brainly.co.id/tugas/21157455
materi tentang pH asam Basa https://brainly.co.id/tugas/21176838
materi tentang pH asam Basa https://brainly.co.id/tugas/21126249
---------------------------------------------------------------------------------------------------
Detail jawaban :
Kelas : 11
Mapel : Kimia
Bab : 5
Kode : 11.7.5
Kata Kunci : asam, basa, pH
10. 1. Calculate the pH and pOH of the solutions given below. (log5 =0.7, log8 = 0.9) a. A solution with a [H + ] of 1x10 -3 M. b. A solution with a [OH - ] of 1x10 -3 M. c. A solution with a [H + ] of 5x10 -7 M. d. A solution with a [OH - ] of 8x10 -11 M.
pH = -log [H⁺]
pOH = -log [OH⁻]
pH = 14-pOH
a) pH = 3-log 10 = 3
pOH = 14-3 = 11
b) pOH = 3-log 10 = 3
pH = 14-3 = 11
c) pH = 7-log 5 = 6,3
pOH = 14-6,3 = 7,7
d) pOH = 11-log 8 = 10,09
pH = 14-10,09 = 3,91
11. an example of a solution between a solid and water is
Jawaban:
Contoh padatan : NaCl(s) ; NaOH(s) ; MgO ; Na₂CO₃ ; KCl ; CaCO₃
Contoh cairan : NH₃(l) , H₂SO₄(l) , HCl(l) , CH₃COOH(l) , HNO₃(l)
SEMOGA MEMBANTU
JADIKAN JAWABAN YERBAIK YA....
(☆^ー^☆)
12. What is the pH of 2×10^-5 M sulphuric acid solution
Jawaban
pH = 4
Pembahasan
Asam kuat
sulphuric acid solution = larutan asam sulfat
larutan asam sulfat = H₂SO₄
H₂SO₄ ---> 2H⁺ + SO₄²⁻
a = 2
[H⁺] = a Ma
[H⁺] = 2 x 2. 10⁻⁵
[H⁺] = 10⁻⁴ M
pH = - log [H⁺]
pH = - log 10⁻⁴
pH = 4
Pelajari lebih lanjut
pH asam kuat dan asam lemah https://brainly.co.id/tugas/732986, https://brainly.co.id/tugas/20903903, https://brainly.co.id/tugas/14852003, https://brainly.co.id/tugas/21082223, https://brainly.co.id/tugas/4965021# pH larutan asam kuat dan basa kuat https://brainly.co.id/tugas/5710167, https://brainly.co.id/tugas/2626244, https://brainly.co.id/tugas/1519985013. An example of a solution between a solid and water is
Jawaban:
Contoh larutan antara zat padat dan air adalah
Penjelasan:maaf jika salah
14. What volume of oxygen reacts with 50cm3 of nitrogen monoxide
Jawaban:
A certain quantity I guess
Penjelasan:
15. 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]
16. in an experiment, 50cm3 of nitric acid solution is mixed with 50cm3 of potassium hydroxide solution in a plastic container. Both solution have the same concertration. when those solution are mixed, the temperature of the mixture increased by 6.0°C. How much heaat is released in the neutralisation reaction? (Spesific heat capacity of water = 4.2 J/g°C, density of water =1.0g/cm3) a. 63.0 J b. 30.0 J c. 25.2 J d. 18.9 J e. 12.6 J
kalau menurut aku jawabannya adalah 25,2 JHNO3 + KOH ---> KNO3 + H2O
dT = 6
mass = 50M mmol
Q = mcdT
Q = 50M.4,2.6
Q = 1260M
17. sodium reacts with water to produce alkaline solution .write the chemical equation
2Na + 2H2O -> 2NaOH + H2
if i'm not mistaken..
18. 400 cm of sodium hydroxide solution 0.5 mol dm 3 is used to neutralise 100 cm of hydrochloric acid solution as shown in the reaction below. NaOH(aq) + HCl(aq) - NaCl(aq) + H2O(1) 4 What is the concentration of the hydrochloric acid solution? A 0.100 mol dm3 С 1.00 mol dm3 B 0.200 mol dm3 D 2.00 mol dm3
Jawaban:
D. 2,00 mol/dm³
Penjelasan:
V NaOH = 400 cm³ = 0,4 dm³
M NaOH = 0,5 mol/dm³ or 0,5 M
V HCl = 100 cm³ = 0,1 dm³
Reaction:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Ask: concentration of HCl?
answer:
neutralise equation is:
Va x Ma x a = Vb x Mb x b
0,1 dm³ x Ma x 1 = 0,4 dm³ x 0,5 M x 1
0,1 dm³ x Ma = 0,2 mol
Ma = 0,2 mol / 0,1 dm³ = 2,00 M or 2,00 mol/dm³
nb:
cm³ = ml
dm³ = L
cmiiw ^^
19. 10 ml of acetic acid solution (Ka=10^-5) with pH = 5 is mixed with 90 ml of water. the pH of the solution now?
this is my answer. please correction again
20. 20.0 cm³ of hydrochloric acid is needed to neutralize completely 25.0 cm³ of 0.1 mol dm-3 sodium hydroxide solution in the conical flask. Calculate the molarity of the hydrochloric acid.
The molarity of the hydrochloric acid is 0,125 M
»EXPLANATION«In this case, you can use 2 ways to solve it: first, stoichiometrically, by first making a balanced reaction and then using moles to calculate molarity (mol/L); and second, which is simpler. Use the formula below:
Va × Ma × a = Vb × Mb × b
where Va is the volume of the acid, Ma is the molarity of the acid, and a is the valence of the acid; the same goes for bases.
Hydrochloric acid (HCl) and sodium hydroxide (NaOH) each have one valence. So, the calculation:
Va × Ma × a = Vb × Mb × b
20.0 cm³ × Ma × 1 = 25.0 cm³ × 0,1 M
20 cm³ × Ma × 1 = 2,5 mmol
Ma = 0,125 M
#GeniuzS05
#SAYIDI
0 komentar:
Posting Komentar