amonia nitrate is prepared commericially by passing ammonia gas through a solution of nitre acid . write down the balnce equation
1. amonia nitrate is prepared commericially by passing ammonia gas through a solution of nitre acid . write down the balnce equation
[STOICHIOMETRY | X SHS]
Ammonium Nitrate : NH₄NO₃
Ammonia : NH₃
Nitrate acid : HNO₃
Reaction :
NH₃ + HNO₃ --> NH₄NO₃ (balanced)
2. ammonia nitrate is prepared commericially by passing ammonia gas througs a solution of nitrate acid .write down the balance equation
[STOICHIOMETRY | X SHS]
Ammonium Nitrate : NH₄NO₃
Ammonia : NH₃
Nitrate acid : HNO₃
NH₃ + HNO₃ --> NH₄NO₃ (balanced)
3. How many grams of silver will be produced if a current of 1.50 A passes through a solution of AgNO3 for 30.0 minutes?
larutan AgNO3 dielektrolisis dengan arus listrik sebesar 1.5 ampere selama 30 menit (1800 detik). Maka reaksi ionisasi AgNO3 adalah :
AgNO3 ⇒ Ag⁺ + NO₃⁻
jika massa atom relatif Ag adalah 107.8 gr/mol maka berat endapan yang terbentuk dikatoda adalah sebanyak :
W = e x i x t /96500
W = (107.8/1)(1.5)(1800)/96500
W = 3.02 gr
jadi jummlah endapan Ag yang terbentuk di elektroda adalah 3.02 gram.
Pembahasan
Sel elektrolisis atau kebalikan sel galvani merupakan sel elektrokimia yang mampu merubah energi listrik menjadi energi kimia. Reaksi kimia tidak spntan dapat berlangsung dengan bantuan muatan listrik pada sel. Prinsip kerja sel volta dalam menghasilkan endapan logam adalah aliran transfer elektron dari reaksi oksidasi di anoda menuju reduksi di katoda melalui rangkain listrik.
Sel elektrolisis secara umum tersusun atas beberapa komponen diantaranya:
anoda merupakan elektrode tempat reaksi oksidasikatoda merupakan elektrode tempat reaksi reduksielektrolit merupakan zat kimia yang mampu menghantarkan arus listrikjembatan garam merupakan rangkain dalam yang merupakan zat garam. Jembatan garam berperan menjaga kesetimbangan ion antara setengah sel anoda dan setengah sel katoda sehingga terbentuk rangkain listrik tertutup.Dalam sel elektrolisis, elektroda yang digunakan bisa berupa elektroda inert seperti karbon (C), Platinum(Pt) dan Emas (Au). Beratnya endapan yang terbentuk (W) setelah sel elektrolisis dijalankan bergantung pada beberapa besaran seperti rumus berikut :
W = e x i x t/96500
dimana i adalah besarnya arus dan t adalah lamanya proses elektrolisis berlangsung.
Pelajari lebih lanjutKorosi di link brainly.co.id/tugas/8144091Persamaan nerst di link brainly.co.id/tugas/7264347Elektrolisis di link brainly.co.id/tugas/9924911Detail tambahan
Kelas : XII SMA
Mapel : Kimia
Materi : Redoks dan Elektrokimia
Kode : 12.7.2
Kata Kunci : Sel Volta, deret potensial reduksi, korosi
4. A solution of H2SO4 with a molal concentration of 8.010 m has adensity of 1.354g/mL. What is the molar concentration of thissolution?
Answer with explanation:
You are given a solution of sulphuric acid with a molal concentration of 8.010 m and its density is 1.354 g/mL.
And we also know that the molar mass of sulphuric acid which is 98.079 g/mol.
◈ ◈ ◈
With the known datas, we work out the molar concentration in mol/L.
◈ ◈ ◈
The relation between molarity, molality and density is described within the following equation.
[tex]\frac{1}{m} = \frac{d}{M}-\frac{Mr}{1000}[/tex]
Where,
Density (d) = 1.354 g/mL
Molecular mass of solute (Mr) = 98.079 g/mol
Molality (m) = 8.010 m
Molarity (M) = ... mol/L
◈ ◈ ◈
So then,
[tex]\frac{1}{8.010\ m} = \frac{1.354\ g/mL}{M}-\frac{98.079\ g/mol}{1000}[/tex]
[tex]0.1248\ m = \frac{1.354\ g/mL}{M}-0.098079\ g/mol[/tex]
[tex]\frac{1.354\ g/mL}{M} = 0.098079\ g/mol+ 0.1248\ m[/tex]
[tex]\frac{1.354\ g/mL}{M} = 0.222879\ g/mmol[/tex]
[tex]M = \frac{1.354\ g/mL}{0.222879\ g/mmol}[/tex]
[tex]M = 6.075\ mol/L[/tex]
◈ ◈ ◈
Therefore, the molar concentration of said acid solution is 6.075 mol/L.
◈ ◈ ◈
~I hope this helps you~
5. 1) The solution of x/2 is...2) The solution of 5x = 2x + 3 is...3) The solution of 4x - 5 = 6x - 1 is...4) The smallest member of the solution set of 3x - 7 ≥ 8 is...5) 4x ≤ 5x + 6 equivalent is...
Jawaban:
2.) 5x = 2x + 3
5x-2x = 3
3x = 3
x= 3
jadi solusi x adalah 3
-----------------------------
3.)
4x-5 = 6x - 1
4x - 6x = 5 - 1
-2x = 4
x = 4: -2
x = -2
---------------------------
4.)
3x - 7 ≥ 8
3x ≥ 8 + 7
3x ≥ 15
x ≥ 15:3
x ≥ 3
Anggota himpunan x adalah {1,2,3}. Jadi anggota terkecil adalah 1 (ITU BILA DIKETAHUI BILANGAN X ADALAH BILANGAN ASLI, KALAU BILANGAN YANG LAIN BISA BERBEDA.)
5.)Mencari persamaan atau kesetaraan dari 4x ≤ 5x + 6
4x ≤ 5x + 6
4x - 5x ≤ 6
-x ≤ 6
Jadi persamaannya juga bisa x ≤ -6
Penjelasan:
1). Mohon maaf saya belum bisa jawab karna saya masih bingung sama soalnya. (Mungkin ada yang kurang?)
2.) Jadi itu suruh mencari solusi alias kita mencari nilai x yaitu dengan mengelompokan bilangan dengan pasangannya terlebih dahulu. (Seperti bilangan ber-Variabel dengan ber-Variabel, dan sebaliknya)
3.)Jadi itu suruh mencari solusi alias kita mencari nilai x yaitu dengan mengelompokan bilangan dengan pasangannya terlebih dahulu. (Seperti bilangan ber-Variabel dengan ber-Variabel, dan sebaliknya)
4.) Jadi kita mencari anggota bilangan yang terkecil dari himpunan x yaitu 1. Ingat ya itu klo diketahui himpunan x adalah bilangan x
5.) Jadi itu tuh suruh mencari kesetaraan atau persamaan. yaitu kita ketahui hasilnya -x ≤ 6. Nah itu bisa disamakan dengan x ≤ 6
Sekian terimakasih, mohon maaf ya klo ada yang keliru.
Semoga membantu! =D.
6. 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 :)
7. What do you observe if ethene is bubbled through bromine water
apa yang anda amati jika etena ditiupkan malaui air brom
]
8. A solution of copper (II) sulphate is Electrlyzed
fgbhgvb
hhfgvb
hgffhbnfcgvc
hhjnnjjElektrolisis CuSO₄ Elektroda Pt
Reaksi di Anoda 2H₂O ==> 4H⁺ + 4e + O₂
Reaksi di Katoda Cu²⁺ + 2e ==> Cu
semoga dapat membantu yah kak
untuk pertanyaan atau apapun kakak bisa mengepostkan di kolom komentar
9. 4) Ammonia gas can be prepared through reaction between nitrogen gas and hydrogen gas. (a) Write the chemical equation for the preparation of ammonia. (b) 600 cm³ of nitrogen gas is mixed with 600 cm³ of hydrogen gas at room conditions. Determine which reactant is used in excess. Calculate the maximum volume of ammonia produced from the reaction. [1 mol of ammonia gas accupies 24 dm³ at room conditions]
Jawaban:
(a) The chemical equation for the preparation of ammonia gas through the reaction between nitrogen gas and hydrogen gas is:N2 + 3H2 -> 2NH3(b) 600 cm³ of nitrogen gas is equal to 600/24 = 25 dm³. Since the molecular weight of nitrogen is 28 g/mol, the mass of 25 dm³ of nitrogen gas is 25 x 28 = 700 g. Similarly, the mass of 600 cm³ of hydrogen gas is 600/24 x 2 = 100 g.
Since the ratio of nitrogen to hydrogen in the reactants is 1:3, the limiting reactant is the one that is present in the lesser amount. In this case, the limiting reactant is hydrogen gas, as it is present in a lesser amount than nitrogen gas. Therefore, the maximum volume of ammonia that can be produced from this reaction is 100/2 = 50 dm³.
10. Exercise 1.2 Rewrite the sentences by putting is or are in the main verbs. 1. One of the glasses broken. 2. Cast-iron not as strong as steel. 3. Oxygen not an inert gas. 4. Oxygen and hydrogen gases. 5. Oxygen, like hydrogen, a gas. 6. This solution a mixture of chlorine and sodium.
Jawaban:
1. One of the glasses is broken.
2. Cast-iron is not as strong as steel.
3. Oxygen is not an inert gas.
4. Oxygen and hydrogen are gases.
5. Oxygen, like hydrogen, is a gas.
6. This solution is a mixture of chlorine and sodium.
Penjelasan:
Semua kalimat yang menggunakan is mempunyai subjek singular/tunggal, dan are mempunyai subjek plural/jamak.
11. what is purpose of problem-solution
In composition, using a problem-solution format is a method for analyzing and writing about a topic by identifying a problem and proposing one or more solutions. A problem-solution essay is a type of argument.
or
to provide a solution to a problem
12. when a dilute solution of salt water is electrolysed a colourless gas is given off from the anode. the gas is?
I think it is oxygen in anode
13. In a solution of saltwater (a saline solution), salt is the?
Salt is the solute.
#ChemistryIsFun
14. When 10 cm³ of 0.5 mol dm3 sodium sulphate solution is added to excess lead (II) nitrate solution, a white precipitate is formed. Calculate the mass of precipitate formed.
Jawaban:
To solve this problem, we need to use the fact that a mole of a substance is a unit of mass equal to the atomic or molecular weight of that substance in grams.
We know that the concentration of the sodium sulphate solution is 0.5 mol/dm³, and that we have added 10 cm³ of the solution. Therefore, the number of moles of sodium sulphate in the solution is 0.5 x (10/1000) = 0.005 moles.
Sodium sulphate has a molecular weight of 142.04 g/mol, so the mass of sodium sulphate in the solution is 0.005 x 142.04 = 0.7102 g.
When sodium sulphate is added to lead (II) nitrate, a white precipitate of lead (II) sulphate is formed. The formula for lead (II) sulphate is PbSO4, and its molecular weight is 303.34 g/mol.
Thus, the mass of lead (II) sulphate precipitate formed is 0.005 x 303.34 = 1.5167 g.
Therefore, the mass of precipitate formed is 1.5167 grams.
15. Water is added to 25.0 ml of a 0.866 m kno3 solution until the volume of the solution is exactly 500 ml. what is the concentration of the fi nal solution?
Jawaban:
Penjelasan:
[tex]M_{KNO_{3}}= 0,866 \ M = \frac{n}{V_{i}} \\\\V_{f}= 500 \ mL\\\\\delta V=25 \ mL\\\\V_{i}=V_{f}-\delta V\\\\V_{i}=500 \ mL-25 \ mL = 475 \ mL\\\\M_{f \ KNO_{3}}=M_{KNO_{3}}\times \frac{V_{i}}{V_{f}} =\\\\M_{f \ KNO_{3}}=0,866 \ M \times \frac{475 \ mL}{500 \ mL} =\\\\M_{f \ KNO_{3}}=0,866 \ M \times \frac{475 \ mL}{500 \ mL} = 0,8227 \ M[/tex]
16. A solution is made by dissolving 25 g of glucose, C₆H₁₂O₆, to make 500 cm³ solution. What is the concentration of the solution in g/dm
Jawaban:
Calculating concentrations
The concentration of a solution can be shown in g/dm3 or mol/dm3. It is often more useful to know the concentration of a reactant in mol/dm3 so that the amount of reactant in a given volume can be calculated.
17. An example of a 0.6025 g of chloride salt was dissolved in water and its chloride was precipitated with excess silver nitrate. The silver chloride precipitate is filtered, washed, dried and weighed 0.7134 g. calculate% Cl in the example
Jawaban:
kita anggap sampelnya adalah XCl
XCl + AgNO3 --> AgCl + XNO3
berat AgCl adalah 0,7134 gram
berat Cl di AgCl = berat Cl di XCl
berat Cl di AgCl = (Ar Ag / Mr AgCl) x berat AgCl
berat Cl di AgCl = (108 / 143,5) x 0,7134
berat Cl di AgCl = 0,53691 gr
% Cl pada sampel = (massa Cl / massa sampel) x 100
% Cl pada sampel = (0,53691 / 0,6025) x 100
% Cl pada sampel = 89,11%
Penjelasan:
:"
18. What is the pH of a 3 x 10-7 M solution of HCl ? What concentration of Hydrogen ions is provided by the ionization of water ?
maaf saya tidak mengerti
19. What is the atomic weight of silver?
Silver nerver and atomic yes.. Never dead is silver the worl go play plK
20. Which of the following statements is correct about chemical bonding in hydrogen gas, H2? a. Hydrogen gas has one covalent bonding with two lone electron pair b. Hydrogen gas has one covalent bonding without lone electron pair c. Hydrogen gas has one covalent bonding with one lone electron pair d. Hydrogen gas has two covalent bonding with one lone electron pair
Jawaban:
lah gw lagi nyari pertanyaan yang sama cuk
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