What is the molarity of a solution containing 4 moles of KCl in 2.5 L of solution?
1. 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 :)
2. What volume is needed to make a 2.45 m solution of kcl using 0.50 mol of kcl?
0,204L
Penjelasan:
Knows :
M = 2.45 molar
n = 0.5 mol
Question :
V..?
Solution :
M = n/V
2.45 = 0,5/V
V = 0,5/2.45
V = 0,204L
so, Volume is needed to make a 2,4M Solution of KCl 0,5mol is 0.204Lor 204mL
Note.
M = Molarity (mole/L or M)
n = mole (mol)
V = volum (L or mL)
3. 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]
4. some useful properties of materials are the deegre of ....... to light ....... property, ...... conducuvity and ....... conductivity
material and sulfur and wild and butter
5. the result of pH of the CH3COONa solution 0,1M
Acid and base
Kw=10^-14
Ka=10^-5
M=0,1M
[H+]=√kw/ka×[M]
=√10^-14/10^-5 ×0,1
=√10^-10
=10^-5 ---> pH=5
6. A solution of salt and water contains 10% salt by weight. 30 kg water evaporates from the solution and now it contains 20% of salt. Find the original quantity of salt.
Please find attached pic of my math work for your question. Hope you can understand.. :)
7. 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.
8. 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
9. In Ca(OH)2 0,05 M solution,the value of pH is
Penjelasan:
Ca(OH)2 adalh basa kuat maka pH = 14 - (2×0.05)= 14 - 0,1 = 13,9
10. find the solution set of |x-2|+1 = 0 a. -6 b.-3 c.0 d. no solution e. 6
|x - 2| = -1
x - 2 = -1
x = 1
atau
-(x - 2) = -1
-x + 2 = -1
3 = x
11. If p = 2 is a solution of the equation 2p - hp +6=0, find the value of h. Hence, find the other solution for the equation.
remember p = 2
Then,
2p² - hp + 6 = 0
2(2)² - h(2) + 6 = 0
2(4) - 2h + 6 = 0
8 + 6 - 2h = 0
14 - 2h = 0
2h = 14
h = 7
so, we have solution h is a 7
12. the solution set of (2x-3)/(x+4)=>0 is
Jawaban:
x≥ -4
Penjelasan dengan langkah-langkah:
(2x-3)/(x+4) ≥0
x + 4 ≥ 0
x ≥ -4
13. 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/1519985014. 4. Find the solution of x² + 8x +16 0!
Jawab:
x² + 8x +16 = 0
(x + 4)^2 = 0
x = - 4
15. The solution of 3(m – 1) + 5 = m + 10 is m =
Jawab:
m = 4
Penjelasan dengan langkah-langkah:
3(m-1) + 5 = m + 10
3m - 3 + 5 = m + 10
3m + 2 = m + 10
3m - m = 10 - 2
2m = 8
m = 4
Hope this helps!
16. 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~
17. 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]
18. 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
19. 22,35 gram of KCL (Mr=74,5) and 22,2 gram of CaCl2 (Mr=111) are dissolved in 400 ml of water. What is the boiling point of the solution?(kb=0,5 derjat celcius kg/mol)
Matter : Colligative Solution
Use this formulation for electrolyte solute
deltaT = m.Kb.i
2 solute as follow:
KCl ----> K+ + Cl- have 2 ions
CaCl2 ---> Ca2+ + 2 Cl- have 3 ions
Find each mol of solute:
KCl: 22.35/74.5 = 0.3 mol
CaCl2: 22.2/111 = 0.2 mol
Both solute dissolved in 400 mL H20
deltaT = 0.3×1000/400×0.5×2 + 0.2×1000/400×0.5×3
= 0.75 + 0.75
= 1.5
So the boiling point of solution become:
100 + 1.5 = 101.5 °C
#ChemistryIsFun
20. Which definition of concentration given below best describes the concept expressed in “in a solution of 10g of salt and 70ml of water, the concentration of the solution by percent mass is 12,5%2?” jawabannya 1.The amount of water in the solution 2 A word denoting 70g 3 The mass of the solution 4 The amount of a component (solute) present. 5 The amount of solution
Jumlah 1. air dalam larutan 2 A 70g kata yang menunjukkan 3 Massa solusi 4 Jumlah komponen ( zat terlarut ) hadir . 5 Jumlah solusi
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