An Li3PO4 solution has an phosphate concentration of 0.910 M. What is the concentration of Li3PO4 solution?
1. An Li3PO4 solution has an phosphate concentration of 0.910 M. What is the concentration of Li3PO4 solution?
Answer:
that's the final answer
2. Van't hoff of Li3PO4?
Answer:
The idea here is that the magnesium hydroxide,
Mg
(
OH
)
2
, will react with the glacial acetic acid,
CH
3
COOH
, to form magnesium acetate,
(
CH
3
COO
)
2
Mg
, and water.
This neutralization reaction will actually consume some of the acetic acid, which means that the resulting solution will contain less solvent than what you start with. Keep this in mind.
The balanced chemical equation for this reaction looks like this
Mg
(
OH
)
2(s]
+
2
CH
3
COOH
(l]
→
2
CH
3
COO
−
+
Mg
2
+
+
2
H
2
O
(l]
Now, the freezing point of a solution depends on the concentration of particles of solute present in solution.
In your case, the particles of solute will include the acetate anion,
CH
3
COO
−
, the magnesium cation,
Mg
2
+
, and water.
In essence, you're looking for any molecule or ion that is not acetic acid.
Mathematically, you can express the freezing-point depression of a solution by using the equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
Δ
T
f
=
i
⋅
K
f
⋅
b
a
a
∣
∣
−−−−−−−−−−−−−−−−−−
, where
Δ
T
f
- the freezing-point depression;
i
- the van't Hoff factor
K
f
- the cryoscopic constant of the solvent;
b
- the molality of the solution.
The van't Hoff factor tells you how many moles of particles of solute you get per mole of solute dissolved in the solvent.
In your case, one mole of magnesium hydroxide will produce
two moles of acetate anions,
2
×
CH
3
COO
−
one mole of magnesium cations,
1
×
Mg
2
+
two moles of water,
2
×
H
2
O
Since you get a total of five moles of particles for every one mole of sodium hydroxide, the van't Hoff factor will be
i
=
5
.
Use the molar masses of magnesium hydroxide and acetic acid (I'll use
HAc
for simplicity) to calculate how many moles of each you're mixing
2.3
g
⋅
1 mole Mg
(
OH
)
2
58.32
g
=
0.03944 moles Mg
(
OH
)
2
25.0
g
⋅
1 mole HAc
60.05
g
=
0.4163 moles HAc
The reaction consumes
2
moles of acetic acid for every
1
mole of magnesium hydroxide, so use this mole ratio to determine how many moles of acetic acid will be consumed by the reaction
0.03944
moles Mg
(
OH
)
2
⋅
2
a
moles HAc
1
mole Mg
(
OH
)
2
=
0.07888 moles HAc
This means that the resulting solution will only contain
n
C
H
3
C
O
O
H
=
0.4163
−
0.07888
=
0.3374 moles HAc
This will be equivalent to
0.3374
moles HAc
⋅
60.05 g
1
mole HAc
=
20.26 g
To get the molality of the solution, use the number of moles of magnesium hydroxide and the mass of acetic acid that remains in the resulting solution.
As you know, to get a solution's molality you need to use number of moles of solute and the mass of the solvent expressed in kilograms.
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
molality
=
number of moles of solute
kilograms of solution
×
100
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, you will have
b
=
0.03944 moles
20.26
⋅
10
−
3
kg
=
1.947 mol kg
−
1
Plug your values into the equation for freezing-point depression
Δ
T
f
=
5
⋅
3.90
∘
C
kg
mol
−
1
⋅
1.947
mol
kg
−
1
Δ
T
f
=
37.97
∘
C
The freezing-point depression is defined as
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
Δ
T
f
=
T
∘
f
−
T
f sol
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−
, where
T
∘
f
- the freezing point of the pure solvent
T
f sol
- the freezing point of the solution
The freezing point of the solution will thus be
T
f sol
=
T
∘
f
−
Δ
T
f
T
f sol
=
16.6
∘
C
−
37.97
∘
C
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
−
21.4
∘
C
a
a
∣
∣
−−−−−−−−−−−−−
I'll leave the answer rounded to three sig figs
Explanation:
Sana naintindihan nyo po
3. NiSO4+Li3PO4->Ni(PO4)2+Li2SO4
Good Day...
NiSO4 + Li3PO4 → Ni3(PO4)2 + Li2SO4
↓ ↓
Reactant Product
Type of reaction: double replacement or double displacement reaction
To balance:
Based on the law of conservation of mass, the mass of reactant is equal to the mass of the product. It also means the number of atoms in each element of reactant and product are equal.
3 NiSO4 + 2 Li3PO4 → Ni3(PO4)2 + 3 Li2SO4
Ni = 3 Ni = 3
SO4 = 3 SO4 = 3
Li = 6 Li = 6
PO4 = 2 PO4 = 2
Hope it helps...=)
4. 1. Given 0.400 mole of Li3PO4, a. how many Li3PO4 units are there? b. how many atoms of P are there?c. how many ions of lithium(Lit) are the? d. how many atoms oxygen are there? e. how many grams of oxygen atoms are there? 2. What is the mass of one mole of CuBr2? 3. What is the mass of 0.84 mole of Al:03?
Answer:
2.Molar mass is
223.37 g/mol
3.26.982 g
Use the periodic table to check the atomic mass, this is the number of grams per mole → 1 mole of Aluminum is 26.982 g
5. Gr.9 po ito pag nasagot nyopo automatic po kayong brainliest I. OBJECTIVE/S: 1. Calculate percentage composition of elements in a compound. 2. Differentiate empirical and molecular formula. 3. Use percentage composition to know the empirical and molecular formula of a compound II. MATERIALS: Periodic Table of Elements, pen, paper, calculator III. PROCEDURES: PART I. Computing Percentage Composition of each Element in a Compound 1. Determine the percentage composition of each atom in calcium carbonate (CaCO3). Look at the periodic table and compute first the molar mass of the compound. a. What is the molar mass of calcium carbonate? ______________ b. What percentage of calcium (Ca) is present in calcium carbonate? _______________ c. What element constitutes the biggest percentage in calcium carbonate? _____________ 2. Determine the mass percentage of each element in lithium phosphate (Li3PO4). Compute first the molar mass of the compound. Round off the percentages to whole number. To check your answer, if you add the mass percentages of all atoms, you should get 100%. a. % Li = _______ % P = _______ %O = _________ b. What element constitutes the biggest percentage in lithium phosphate?
Calcium Carbonate (CaCO3)
A. What is the molar mass of calcium carbonate?
Answer: Molar Mass = 100.11g/mol
Note: To get the molar mass of a compound, you must get the mass of each element presented in the given compound and add them to get the total mass.
Ca = 40.1g/mol
C = 12.01g/mol
O = 16g/mol × 3 = 48g/mol
Molar Mass:
= 40.1g/mol + 12.01g/mol + 48g/mol
= 100.11g/mol
Note: Oxygen is multipled by 3, because the compound consist of 3 oxygen.
B. What percentage of calcium (Ca) is present in calcium carbonate?
Answer: %Ca = 40%
Note: To get the percentage composition of an element in a compound, use this formula below;
Mass of Element / Molar Mass × 100%
Molar Mass = 100.11g/mol
Ca = 40.1g/mol
Percentage Composition of Calcium:
= 40.1g/mol / 100.11g/mol × 100%
= 40%
C. What element constitutes the biggest percentage in calcium carbonate?
Answer: Oxygen
Note:
Percentage Composition of Carbon:
= 12.01g/mol / 100.11g/mol × 100%
= 12%
Percentage Composition of Oxygen:
= 48g/mol / 100.11g/mol × 100%
= 48%
Lithium Phosphate (Li3PO4)A. %Li = 18%
%P = 27%
%O = 55%
Note: First step, get the molar mass of the compound first, to get the percentage composition of each element.
Li = 7g/mol × 3 = 21g/mol
P = 31g/mol
O = 16g/mol × 4 = 64g/mol
Molar Mass
= 21g/mol + 31g/mol + 64g/mol
= 116g/mol
Note: Now we have the molar mass, let’s compute the percentage composition of each element in the given compound.
Formula: Mass / Molar Mass × 100%
Percentage Composition of Lithium:
= 21g/mol / 116g/mol × 100%
= 18%
Percentage Composition of Phosphorus:
= 31g/mol / 116g/mol × 100%
= 27%
Percentage Composition of Oxygen:
= 64g/mol / 116g/mol × 100%
= 55%
Note: To check if your answer is right, get the total percentage by adding them up and the answer must always be equal to 100%.
18% + 27% + 55% = 100%
B. What element constitutes the biggest percentage in lithium phosphate?
Answer: Oxygen
P.S. Please correct me if i’m wrong tyia
ฅ^•ﻌ•^ฅ
6. What mass of phosphoric acid, H3PO4, would actually react with 7.17 grams of LiOH? 3LiOH + H3PO4 ® Li3PO4 + 3H2O
SOLUTION:
Step 1: Calculate the molar mass of H₃PO₄ and LiOH.
The molar masses of H, P, O, and Li are 1.008 g, 30.974 g, 15.999 g, and 6.94 g, respectively.
• For H₃PO₄
[tex]\begin{aligned} MM_{\text{H}_3\text{PO}_4} & = \text{3(1.008 g) + 30.974 g + 4(15.999 g)} \\ & = \text{97.994 g} \end{aligned}[/tex]
• For LiOH
[tex]\begin{aligned} MM_{\text{LiOH}} & = \text{6.94 g + 15.999 g + 1.008 g} \\ & = \text{23.95 g} \end{aligned}[/tex]
Step 2: Calculate the mass of H₃PO₄ reacted with LiOH.
Based on the balanced chemical equation, 3 moles of LiOH is stoichiometrically equivalent to 1 mole of H₃PO₄.
[tex]\begin{aligned} \text{mass of} \: \text{H}_3\text{PO}_4 & = \text{7.17 g LiOH} \times \frac{\text{1 mol LiOH}}{\text{23.95 g LiOH}} \times \frac{\text{1 mol} \: \text{H}_3\text{PO}_4}{\text{3 mol LiOH}} \times \frac{\text{97.994 g} \: \text{H}_3\text{PO}_4}{\text{1 mol} \: \text{H}_3\text{PO}_4} \\ & = \boxed{\text{9.78 g}} \end{aligned}[/tex]
Hence, 9.78 g of H₃PO₄ would actually react with 7.17 g of LiOH.
7. II. Balance the following chemical equations. 1. P4 + O2 → P2O5 2. CH4 + O2 CO2 + 2 H₂O 3. Fe + NaBr →>>> FeBr3+ Na 4. CaSO4 + Mg(OH)2 Ca(OH)2 + MgSO4 5. NH4OH + HBr → H₂O + NH4Br 6. P4 + O2 P₂O5 7. NaNO3 NaNO2 + O2 8. C18H18 + O2 → CO2 + H₂O 9. H₂SO4 + NaOH → NaSO4 + H₂O 10. NiSO4 + Li3PO4 → Ni3 (PO4)2 11. Zn + HCl → ZnCl₂ + H₂ 12. CH4 + O2 CO2 + H₂O FeBr3 + Na 13. Fe + NaBr →FeBr3 + Na 14. SiCl4 + H₂O → SiO₂ + HCI 15. N2 + O2 + H₂O → HNO3
answer is in the pic
sorry i don't know 11 to 15
Explanation:
꧁᪥✵✯ᕼOᑭᗴ IT ᕼᗴᒪᑭՏ✯✵᪥꧂
8. . Determine the mass percentage of each element in lithium phosphate (Li3PO4). Compute first the molar mass of the compound. Round off the percentages to whole number. To check your answer, if you add the mass percentages of all atoms, you should get 100%. a. % Li = _______ % P = _______ %O = _________ b. What element constitutes the biggest percentage in lithium phosphate?
Answer:
wala akong alam sa lesson mo pero hope you feel better and wag ka pastress
9. Classify the following unbalanced chemical equations according to six types of chemical reactions. A. Combination B. Decomposition C. Single Displacement D. Double Displacement E. Combustion F. Acid-base 1. NaOH + KNO3 > NaNO3 + KOH 2. CH4 + O2 > CO2 + 2 H2O 3. Fe + NaBr > FeBr + Na 4. CaSO + Mg(OH)2 > Ca(OH)2 + MgSO4 5. NH4OH + HBr > H2O + NH4Br 6. P4 + O2 > P2O5 7. NaNO > NaNO2 + O2 8. C18H18 + O2 > CO2 + H2O 9. H2SO4 + NaOH > NaSO4 + H2O 10. NiSO4 + Li3PO4 > Ni3(PO4)2 + Li2SO4 ASAP PLEASE. THANK YOU SO MUCH. SO MUCH LOVE. MWAH :*
1. NaOH + KNO₃> NaNO₃ + KOH : Double Displacement
2. CH₄ + O₂> CO₂ + 2 H₂O : Combustion
3. Fe + NaBr--> FeBr + Na : Single Displacement
4. CaSO₄ + Mg (OH)₂--> Ca (OH)₂ + MgSO₄ : Double Displacement
5. NH₄OH + HBr--> H₂O + NH₄Br : Acid-base
6. P₄ + O₂---> P₂O₅ : Combination
7. NaNO₃> NaNO₂ + O₂ : Decomposition
8. C₈H₁₈ + O₂--> CO₂ + H₂O : Combustion
9. H₂SO₄ + NaOH--> NaSO₄ + H₂O : Acid-base
10. NiSO₄ + Li₃PO₄> Ni₃ (PO4)₂+ Li₂SO₄ : Double Displacement
Further explanationThere are chemical reactions that may occur.
1. A single replacement reaction is a chemical reaction in which one element replaces the other elements of a compound to produce new elements and compoundsNot all of these reactions can occur. We can use the activity series, which is a list of elements that can replace other elements below / to the right of them in a single replacement reaction.
This series is better known as the Volta series, where the metal element with a more negative electrode potential is on the left, while the element with a more positive electrode potential on the right.
The more left the position of a metal in the series, the more reactive metal (easy to release electrons, the stronger the reductor)
Generally, the Volta series used is
Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb Bi Cu Hg Ag Pt Au
Example:
2Al (s) + 3Zn (NO₃) ₂ (aq) → 2Al (NO₃) ₃ (aq) + 3Zn (s)
The existing aluminum element can replace the zinc element, which is on the right side so the reaction will occur.
2. Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the productTo predict whether this reaction can occur or not is one of them, the precipitation reaction. A precipitation reaction occurs if two ionic compounds which are dissolved reacted to produce one of the products of the ion compound does not dissolve. Formation of these precipitating compounds that cause reactions can occur
Solubility Rules:
• 1. soluble compound
All compounds of Li +, Na +, K +, Rb +, Cs +, and NH4 +
All compounds of NO₃⁻ and C₂H₃O₂⁻
Compounds of Cl−, Br−, I− except Ag⁺, Hg₂²⁺, Pb²⁺
Compounds of SO₄²⁻ except Hg₂²⁺, Pb²⁺, Sr²⁺, Ba²⁺
• 2. insoluble compounds
Compounds of CO₃²⁻ and PO₄³⁻ except for Compounds of Li +, Na +, K +, Rb +, Cs +, and NH₄ +
Compounds of OH− except Compounds of Li +, Na +, K +, Rb +, Cs +, NH₄⁺, Sr²⁺, and Ba²⁺
3. Combination reactionsA combination of two or more elements or compounds to form a single compound
4. Combustion reaction: The reaction of a substance with oxygenPerfect combustion of carbon compounds will get CO2 gas, whereas if it is not perfect it will produce CO gas
5. Acid-base reaction: the reaction between acid and base to produce salt
1. NaOH + KNO₃> NaNO₃ + KOH
The reaction between the NaOH base and KNO₃ salt, including Double Displacement
2. CH₄ + O₂> CO₂ + 2 H₂O
Methane reaction with O₂ including Combustion reactions
3. Fe + NaBr--> FeBr + Na
The reaction to replace Fe with Na is included as Single Displacement, but this reaction will not occur because the Fe's reactivity is lower than Na
4. CaSO₄ + Mg (OH)₂--> Ca (OH)₂ + MgSO₄
The reaction of CaSO₄ salt and base Mg (OH)₂ includes Double Displacement
5. NH₄OH + HBr--> H₂O + NH₄Br
The weak base reactions of NH₄OH and strong acids HBr include Acid-base reaction
6. P₄ + O₂---> P₂O₅
The phosphorus and O₂ reactions are included in the Combination reaction
7. NaNO₃> NaNO₂ + O₂
This reaction includes decomposition because NaNO₃ breaks down into NaNO₂ + O₂
8. C₈H₁₈ + O₂--> CO₂ + H₂O
Hydrocarbon and O₂ reactions are combustion reactions which produce CO₂ and H₂O gas
9. H₂SO₄ + NaOH--> NaSO₄ + H₂O
Strong acid H₂SO₄ reaction and strong base NaOH including acid base reaction
10. NiSO₄ + Li₃PO₄> Ni₃ (PO4)₂+ Li₂SO₄
This reaction includes Double Displacement because the two component elements replace each other in forming new compounds in the product
Learn more
the symbol for a chemical reaction
https://brainly.ph/question/2122300
statements describes a double replacement reaction
https://brainly.ph/question/2155865
Balancing chemical reaction
https://brainly.ph/question/532985
