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
1. 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.
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 solution of 0.18 g of a non-electrolyte substance is dissolved in 2.5 g of water. If the freezing point of the solution is -1.34 c, the molar mass of the substance is ... (k, water= 1.86 c kg\mol
m a = 0,18 gram (a = non-electrolyte substance)
p = 2,5 gram = 2,5/1000 = 0,0025 kg (p = mass of water)
kf = 1,86°C kg/mol (kf = k water)
Tf a = -1,34°C (Tf = freezing point)
Tf b = 0°C (freezing point of water is 0°C)
Mr = ? (Mr = molar mass of the substance)
answer :
Δ Tf = Tf b - Tf a (Δ Tf = reduction of the freezing point)
= 0 - (-1,34)
= 1,34°C
ΔTf = kf × m (m = molality)
1,34 = 1,86 × m
m = 1,34/1,86 = 0,72 kg/mol
m = n/p (n = mol of the substance)
0,72 = n/0,0025
n = 0,72 × 0,0025 = 0,0018 mol
n = m a/Mr
0,0018 = 0,18/Mr
0,0018Mr = 0,18
Mr = 0,18/0,0018 = 100 gram/mol
so, the molar mass of the substance is 100 gram/mol
sorry if the answer is wrong
4. 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
5. If 149.5 g of potassium chloride is dissolved in water to make 500 ml solution, what is the molarity and in g/dm3 of the solution?
Jawaban:
[tex]=\frac{149.5}{74.55}x\frac{1000}{500} \\=4.01 molar[/tex]
Penjelasan:
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. 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]
8. 15. At temperature 63.5 °C, the vapour pressures of pure H20 and pure ethanol, C2H5OH,are 175 and 4.00 x 102 torr, respectively. A solution is prepared by mixing equal masses ofH2O and C2H5OH.a) Calculate the mole fraction of ethanol in the solution.b) What is the vapour pressure of the solution at 63.5 °C?c) Determine the mole fraction of ethanol in the vapour phase above the solution.
Jawaban:
SELAM
Penjelasan:
Soru basitti
Cevap
merıpbdvre m
BAŞARILAR
9. 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)
10. One solution of the quadratic equation x^2 + px + 12=0 is 3, then p=... and another solution is...
x² + px + 12 = 0
salah astu penyelesaian x = 3
maka
x² + px + 12 = 0
⇒ (3)² + p(3) + 12 = 0
⇒ 9 + 3p + 12 = 0
⇒ 3p = -21
⇒ p = -7
Sehingga
x² - 7x + 12 = 0
⇒ (x - 3)(x - 4) = 0
maka
x - 3 = 0 ⇒ x = 3
x - 4 = 0 ⇒ x = 4
Jadi, p = -7, dan penyelesaian yang satunya = 4
Terimakasih semoga membantu
11. What is the molarity of H2SO4 solution which contains 49.04 g in 1 litre solution.
Materi : Asam - Basa
Answer:
0.500408163236 (0.5 molarity)
Explanation:
to calculate the molarity of sulfuric acid H2SO4.
KNOWN;
m H2SO4 = 49.04 g
v = 1 litre
Mr H2SO4 =
Mr H2SO4
= (2 × Ar h) + (1 × Ar S) + (4 x Ar o)
= (2×1) + (1×32) + (4×16)
= 2+32+6
Mr H2SO4 = 98
(1.The formula for calculating the moles of sulfuric acid
n = m/Mr H2SO4
n = 49.04/98
= 0.50040816326 or 0.5
(2.The formula for calculating the molarity of sulfuric acid
M = n/v
M = 0.50040816336/1
= 0.5
→so the molarity of sulfuric acid H2SO4 is 0.5
hope it helps, thank you
12. If 0 < a < 1, so the solution set of inequality a is...
If 0 < a < 1, the solution set of inequality a is all real numbers that are greater than 0 and less than 1.
In other words, the solution set of the inequality a is the interval (0,1), which includes all real numbers x such that 0 < x < 1.
For example, if a = 0.5, then the solution set of the inequality a is the interval (0,1), which includes all real numbers x such that 0 < x < 1. This interval includes all numbers between 0 and 1, including 0.5.
If a is any other number between 0 and 1, the solution set of the inequality a will also be the interval (0,1).
I hope this helps!
13. The molarity of solution that is made from 6.73 g Na2CO3 which is dissolved in enough water to make 250 mL of solution is …. (Ar Na = 23; C = 12; O = 16)
Jawaban:
Mr Na2CO3 = 2xArNa + 1xArC + 3×ArO
= 2x23 + 1x12 + 3x16
= 46 + 12 + 48
= 106
mol = massa/Mr
= 6,73 g/106
= 0,0635
molarity = mol/V
= 0,0635/0,25 liter
= 0,254
14. 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)
15. In a solution of saltwater (a saline solution), salt is the?
Salt is the solute.
#ChemistryIsFun
16. PRACTICE EXAMPLE B: A 11.3 mL sample of CH3OH (d = 0.793 g/mL) is dissolved in enough water to produce 75.0 mL of a solution with a density of 0.980 g/mL. What is the solution concentration expressed as (a) mole fraction H20; (b) molarity of CH2OH; (c) molality of CH4OH?
Jawaban:
Take a 1.0 L volume of the solution
This will have mass = 980 g
It will contain 11.3 mL/ 75 mL solution * 1000 mL/L solution = 150.67 mL CH3OH
Mass of CH3OH = 150.67 mL * 0.793 g/mL = 119.5 g
Solution is therefore 119.5 g CH3OH + (980 g - 119.5 g) = 860.5 g wtaer
molar mass CH3OH = 32 g/mol
Mol CH3OH = 119.5 g / 32 g/mol = 3.73 mol in 860.5 g H20
Mol in 1000 g H20 = 1000 g H20 / 860.5 g H20 * 3.73 mol = 4.33 mol in 1.0 kg water
Molality = 4.33 m
Calculate mol fraction :
CH3OH = 4.33 mol
H20 = 1000 g / 18 g/mol = 55.55 mol
Total moles = 59.88 mol
mol fraction CH3OH = 4.33 mol / 59.88 mol = 0.072
Mol fraction H20 = 55.55 mol / 59.88 mol = 0.928
mol fraction has no units. Sum of mol fractions = 1.0
Penjelasan:
SEMOGA MEMBANTU MU
SEMOGA MEMBANTU MUJADIKAN JAWABAN SAYA SEBAGAI TERBAIK DAN TERCERDAS
17. 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
18. 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.
19. 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
20. 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
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