[tex]FeCl_3[/tex] is added to ensure all reducing agent is oxidized, indicating endpoint in dichromate titration.
In a dichromate titration, [tex]FeCl_3[/tex] is frequently added to the example arrangement as a sign of the endpoint. The expansion of [tex]FeCl_3[/tex] to the example arrangement assists with guaranteeing that the lessening specialist has been all oxidized by the potassium dichromate ([tex]K_2Cr_2O_7[/tex] ) arrangement.
[tex]FeCl_3[/tex] responds with any overabundance[tex]K_2Cr_2O_7[/tex] in the answer for structure a red-earthy colored encourage of[tex]Fe(OH)_3[/tex] , demonstrating that the lessening specialist has been all oxidized. This response is known as a "back-titration" since overabundance[tex]K_2Cr_2O_7[/tex] is added to the arrangement, trailed by the option of [tex]FeCl_3[/tex] to decide how much unreacted [tex]K_2Cr_2O_7[/tex] .
The response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] can be addressed as:
[tex]6FeCl_3 + K_2Cr_2O_7 + 7H_2SO_4 → 3Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + K_2SO_4 + 7H_2O + 3Cl_2[/tex]
The [tex]FeCl_3[/tex] goes about as a pointer in this response since it responds with the overabundance [tex]K_2Cr_2O_7[/tex] until the decreasing specialist has been all oxidized. As of now, the red-earthy colored encourage of [tex]Fe(OH)_3[/tex] structures, showing that the endpoint has been reached.
Without the expansion of [tex]FeCl_3[/tex], it would be challenging to precisely decide the endpoint of the titration. The expansion of [tex]FeCl_3[/tex] is important to guarantee that the lessening specialist has been all oxidized and that the endpoint has been reached, considering a more exact assurance of the grouping of the diminishing specialist in the example arrangement.
In outline, the expansion of [tex]FeCl_3[/tex] to the example arrangement in a dichromate titration is significant in light of the fact that it assists with guaranteeing that the decreasing specialist has been all oxidized, considering a more precise assurance of its focus.
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The complete question is -
What is the reason for a dichromate titration, and how does [tex]FeCl_3[/tex]support deciding the endpoint of the titration? Might you at any point give the compound condition to the response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] and make sense of why [tex]FeCl_3[/tex] goes about as a marker in this response?
How much aluminum can be produced from 9.00 ton of Al2O3?
To calculate the amount of aluminum produced from 9.00 tons of Al2O3, we need to use stoichiometry. First, we'll convert the mass of Al2O3 to moles, and then use the balanced chemical equation to find the moles of aluminum. Finally, we'll convert the moles of aluminum back to mass.
1. Convert mass of Al2O3 to moles:
9.00 tons = 9,000 kg
Molar mass of Al2O3 = (2 * 26.98) + (3 * 16.00) = 101.96 g/mol
9,000 kg * (1000 g/kg) = 9,000,000 g
moles of Al2O3 = 9,000,000 g / 101.96 g/mol = 88,258 moles
2. Use balanced chemical equation to find moles of aluminum:
The balanced chemical equation is:
2 Al2O3 → 4 Al + 3 O2
Using stoichiometry, we find the ratio of Al2O3 to Al is 2:4 or 1:2.
moles of Al = 88,258 moles Al2O3 * (2 moles Al / 1 mole Al2O3) = 176,516 moles
3. Convert moles of aluminum back to mass:
Molar mass of Al = 26.98 g/mol
Mass of Al = 176,516 moles * 26.98 g/mol = 4,762,984 g
Mass of Al in tons = 4,762,984 g / (1000 g/kg) / (1000 kg/ton) = 4.76 tons
So, 4.76 tons of aluminum can be produced from 9.00 tons of Al2O3.
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Gaseous butane (CH3(CH2)2CH3) will react with gaseous oxygen (02) to produce carbon dioxide (CO2) and gaseous water (H2O). Suppose 34.g of butane s mixed with 200. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.
Determining the maximum mass of water producedThe balanced chemical equation for the reaction between butane and oxygen is:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.
moles of butane = 34. g / 58.12 g/mol = 0.585 mol
moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol
Determining the limiting reactant.
butane : oxygen = 0.585 mol : 6.25 mol
= 0.0936 : 1.00
stoichiometric ratio = 1 : 13/2
= 0.7692 : 1.00
Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.
The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).
moles of water = 5/13 * 6.25 mol
= 2.403 mol
Finally, we can convert the moles of water to grams:
mass of water = 2.403 mol * 18.015 g/mol
= 43.3 g
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A gas‑filled weather balloon has a volume of 56.0 L
at ground level, where the pressure is 761 mmHg
and the temperature is 23.1 ∘C.
After being released, the balloon rises to an altitude where the temperature is −6.97 ∘C
and the pressure is 0.0772 atm.
What is the weather balloon's volume at the higher altitude?
A gas‑filled weather balloon has a volume of 56.0 L
at ground level, where the pressure is 761 mmHg
and the temperature is 23.1 ∘C.
After being released, the balloon rises to an altitude where the temperature is −6.97 ∘C
and the pressure is 0.0772 atm.
What is the weather balloon's volume at the higher altitude?
We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:
(P1 x V1) / T1 = (P2 x V2) / T2
where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.
We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:
(P1 x V1) / T1 = (P2 x V2) / T2
V2 = (P1 x V1 x T2) / (P2 x T1)
V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)
V2 = 2,040 L (rounded to three significant figures)
Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.
Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6 , and an unknown amount of propane, C3H8 ) were added to the same 10.0- L container. At 23.0 ∘C, the total pressure in the container is 3.70 atm. Calculate the partial pressure of each gas in the container.
The partial pressure of each gas are:
Partial pressure of CH₄ is 1.22 atmPartial pressure of C₂H₆ is 1.46 atmPartial pressure of C₃H₈ is 1.02 atmHow do i determine the partial pressure of each gas?First, we shall determine the mole of 8.00 g of methane, CH₄ and 18.0 g of ethane, C₂H₆. Details below:
For methane, CH₄
Mass of CH₄ = 8 g Molar mass of CH₄ = 16 g/mol Mole of CH₄ =?Mole = mass / molar mass
Mole of CH₄ = 8 / 16
Mole of CH₄ = 0.5 mole
For ethane, C₂H₆
Mass of C₂H₆ = 18 g Molar mass of C₂H₆ = 30 g/mol Mole of C₂H₆ =?Mole = mass / molar mass
Mole of C₂H₆ = 18 / 30
Mole of C₂H₆ = 0.6 mole
Next, we shall determine the total mole. Details below:
Volume (V) = 750 mL = 10 LTemperature (T) = 23 °C = 23 + 273 = 296 KPressure (P) = 3.70Gas constant (R) = 0.0821 atm.L/mol KTotal of mole (n) =?PV = nRT
3.70 × 10 = n × 0.0821 × 293
Divide both sides by (0.0821 × 293)
n = (3.70 × 10) / (0.0821 × 293)
n = 1.52 mole
Finally, we shall determine the partial pressure of each gas. Details below:
For methane, CH₄
Mole of CH₄ = 0.5 moleTotal mole = 1.52 moleTotal pressure = 3.70 atmPartial pressure of CH₄ =?Partial pressure = (Mole / total mole) × total pressure
Partial pressure of CH₄ = (0.5 / 1.52) × 3.70
Partial pressure of CH₄ = 1.22 atm
For ethane, C₂H₆
Mole of C₂H₆ = 0.6 moleTotal mole = 1.52 moleTotal pressure = 3.70 atmPartial pressure of C₂H₆ =?Partial pressure = (Mole / total mole) × total pressure
Partial pressure of C₂H₆ = (0.6 / 1.52) × 3.70
Partial pressure of C₂H₆ = 1.46 atm
For propane, C₃H₈
Partial pressure of CH₄ = 1.22 atmPartial pressure of C₂H₆ = 1.46 atmTotal pressure = 3.70 atmPartial pressure of C₃H₈ =?Partial pressure of C₃H₈ = Total pressure - (Partial pressure of CH₄ + Partial pressure of C₂H₆)
Partial pressure of C₃H₈ = 3.7 - (1.22 + 1.46)
Partial pressure of C₃H₈ = 1.02 atm
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what is the ph of a .100M naclo solution
The pH of a 0.100M NaClO solution is 1.
How to calculate pH?pH, meaning power of hydrogen, is a measure of how acidic/basic a solution is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.
pH is really a measure of the relative amount of free hydrogen and hydroxyl ions in the water. It can be estimated using the following formula;
pH = - log {H+}
Where;
H+ = hydrogen ion concentrationpH = - log {0.100}
pH = 1
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An unknown alkene is ozonolyzed and worked up under oxidizing conditions. The H NMR spectrum of the only product obtained is shown. Identify the alkene.
To identify the unknown alkene based on its H NMR spectrum, a qualified organic chemist would need to analyze the chemical shifts, integration values, and splitting patterns of the peaks in the spectrum, and compare them with known reference data and other spectroscopic techniques (such as C NMR, IR, and mass spectrometry) to make an accurate determination.
The alkene is likely to be a symmetrical alkene with two equivalent methyl groups attached to the double bond. This can be seen from the singlet at 1.7 ppm, which is characteristic of a methyl group, appearing twice in the spectrum. The ozonolysis of the alkene would lead to the formation of two carbonyl compounds, which are then oxidized to carboxylic acids under the given oxidizing conditions. Therefore, the alkene in question is likely to be cis-2-butene.
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Three of the primary components of air are
carbon dioxide, nitrogen, and oxygen. In a
sample containing a mixture of only these
gases at exactly one-atmosphere pressure, the partial pressures of carbon dioxide and nitrogen are given as PCO2 = 0.285 torr and
PN2 = 580.502 torr. What is the partial pressure of oxygen?
Answer in units of torr.
The partial pressure of the oxygen is 0.236 atm.
What is partial pressure?The pressure that one gas component in a mixture of gases exerts is known as partial pressure. It is the pressure that the gas would experience if it took up the same amount of space in the mixture at the same temperature on its own.
We know that;
P[tex]CO_{2}[/tex] = 0.285 torr or 0.000375 atm
P[tex]N_{2}[/tex] = 580.502 torr or 0.764 atm
P[tex]O_{2}[/tex] = ?
Total pressure = 1 atm
Then we have that;
PT =P[tex]CO_{2}[/tex] +P[tex]N_{2}[/tex]+ P[tex]O_{2}[/tex]
P[tex]O_{2}[/tex] = PT - (P[tex]CO_{2}[/tex] + P[tex]N_{2}[/tex])
P[tex]O_{2}[/tex] = 1 - (0.000375 + 0.764)
P[tex]O_{2}[/tex]= 0.236 atm
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What is the equilibrium constant, K? 3 A(g) + 3 B(g) <-> 5 C(g) + 2 D(g)
The equilibrium constant is written as;
Keq = [tex][D]^2 [C]^5/[A] [B]^3[/tex]
What is the equilibrium constant?The equilibrium constant's value is influenced by the reaction's chemical make-up and temperature.
The product of the product concentrations, each raised to the power of their stoichiometric coefficient, divided by the product of the reactant concentrations, each raised to the power of their stoichiometric coefficient, is known as the equilibrium constant.
The equilibrium constant is Keq = [tex][D]^2 [C]^5/[A] [B]^3.[/tex]
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help with questions 1-5 pls??
In comparison to towns located inland, cities close to water features like lakes or oceans typically experience cooler summer temperatures.
Why is a city not so hot in summer when the city is close to water?Since water has a higher specific heat capacity than land, this is the case. The quantity of energy needed to raise a substance's temperature by a specific amount is known as its specific heat capacity. Compared to land, raising the temperature of water requires more energy because water has a higher specific heat capacity.
The summer sun warms both land and water, but due to land's lower specific heat capacity, land warms up more quickly than water. As a result, communities farther from water bodies tend to be hotter than cities closer to water.
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Five types begging the question
Five types of begging the question include: Circular reasoning, Loaded question, False analogy, Suppressed evidence and Appeal to authority.
Begging the question is a logical fallacy that occurs when someone assumes the truth of a premise in their argument, without providing evidence or proof. There are several types of begging the question:
1. Circular reasoning: This occurs when someone uses their conclusion as one of their premises, essentially assuming what they are trying to prove.
Example: "God exists because the Bible says so, and the Bible is the word of God."
2. Loaded question: This occurs when someone asks a question that assumes a particular answer or perspective.
Example: "Have you stopped beating your spouse yet?" This question assumes that the person being asked was previously beating their spouse.
3. False analogy: This occurs when someone uses an analogy that is not relevant or applicable to the argument at hand.
Example: "Banning guns is like banning cars because both can be used to kill people." This analogy is false because cars have a primary function of transportation, whereas guns have a primary function of killing.
4. Suppressed evidence: This occurs when someone ignores or dismisses evidence that contradicts their argument.
Example: "I don't believe in climate change because it's cold outside today." This argument suppresses evidence that shows long-term trends of warming temperatures.
5. Appeal to authority: This occurs when someone uses an authority figure or expert as evidence, without providing any other support for their argument.
Example: "Dr. Smith says that this diet is the best for losing weight, so it must be true." This argument appeals to Dr. Smith's authority without providing any evidence or research to support the claim.
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What happens to a buffered solution when a small amount of base is added?
O The solution quickly becomes neutral.
O The solutions resists changes in pH.
O The solution slowly becomes acidic.
O The solution quickly becomes basic.
Answer:
solution resists changes in pH
Explanation:
the inherent property of buffers is to resist change to ph even when acids and bases are added. when the base is added, it is quickly neutralized by the conjugate acid, so the ph won't change.
Answer:
B: The solutions resists changes in pH.
Explanation:
Buffer reactions maintain stable pH of solutions.
Macmillan Learning Determine the formal charge on each atom in the structure. H H-B-H H What is the overall charge on the structure? -2 +1 Answer Bank +2 +3 -3 -4 +4 0
The overall charge on the structure is negative one (-1).
The central boron atom in the structure is bonded to two hydrogen atoms. Boron has three valence electrons, and it has formed only two bonds, so it has a formal charge of +1.
Each of the hydrogen atoms has one valence electron, and each is bonded to the boron atom, so each hydrogen atom has a formal charge of -1. The sum of the formal charges in the structure is equal to the charge of the ion, which is -2. Adding up the formal charges of the atoms, we get:
B: +1
H: -1 (two times)
Overall charge = sum of formal charges = +1 - 1 - 1 = -1
Therefore, the overall charge on the structure is negative one (-1).
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A specific organic reaction is described by the energy diagram drawn below. Using this
energy diagram, identify which product will form first and which product will be the
major product if given enough time?
The product that is formed first is product B and the product that will be the major product if given enough time would also be product B.
In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products.
The activation energy of the reaction can be shown on a diagram as the energy between the reactants that the transition state.
The product with lesser activation energy is the product that is formed first and the major product is decided by the stability of the product which depends on the energy. Lesser is the energy of the product, greater is its stability.
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A sample of an ideal gas has a volume of 2.31 L
at 279 K
and 1.01 atm.
Calculate the pressure when the volume is 1.09 L
and the temperature is 308 K.
The pressure of the gas when the volume is 1.09 L and the temperature is 308 K is 2.36 atm.
What is the final pressure of the gas?The final pressure of the gas is calculated by applying ideal gas law as follows;
(P₁V₁)/T₁ = (P₂V₂)/T₂
where
P₁, V₁, and T₁ are the initial pressure, volume, and temperature, P₂, V₂, and T₂ are the final pressure, volume, and temperature,P₂ = (P₁V₁ x T₂)/(V₂ x T₁)
P₂ = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)
P₂ = 2.36 atm
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1. Which metal is the most reactive? How do you know this?
2. Rank the metals in order of increasing reactivity.
3. Give the chemical equations for each single replacement reaction that took place.
4. Was Fe^3+ reduced? Of so what metal(s) acted as reducing agents?
Please help thanks so much!!!!!!!!!!!!
The total mass of products obtained when 130 g of zinc react completely with HCl is 274 g (3rd option)
How do i determine the total mass of products obtained?First, we shall determine the mass of each product obtained. Details below:
For ZnCl₂
2HCl + Zn -> ZnCl₂ + H₂
Molar mass of Zn = 65 g/molMass of Zn from the balanced equation = 1 × 65 = 65 g Molar mass of ZnCl₂ = 135 g/molMass of ZnCl₂ from the balanced equation = 1 × 135 = 135 gFrom the balanced equation above,
65 g of Zn reacted to produce 135 g of ZnCl₂
Therefore,
130 g of Zn will react to produce = (130 × 135) / 65 = 270 g of ZnCl₂
Thus, the mass of ZnCl₂ obtained is 270 g
For H₂
2HCl + Zn -> ZnCl₂ + H₂
Molar mass of Zn = 65 g/molMass of Zn from the balanced equation = 1 × 65 = 65 g Molar mass of H₂ = 2 g/molMass of H₂ from the balanced equation = 1 × 2 = 2 gFrom the balanced equation above,
65 g of Zn reacted to produce 2 g of H₂
Therefore,
130 g of Zn will react to produce = (130 × 2) / 65 = 4 g of H₂
Thus, the mass of H₂ obtained is 4 g
Finally, we shall determine the total mass of the product produced. Details below:
Mass of ZnCl₂ = 270 gMass of H₂ = 4 gTotal mass of product =?Total mass of product = mass of ZnCl₂ + mass of H₂
Total mass of product = 270 + 4
Total mass of product = 274 g (3rd option)
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diffrences in water temperature in the ocean create movement because-
Diffrences in water temperature in the ocean create movement because bodies of water at different temperatures have different densities.
How can the differences be explained?Water that is colder is generally denser than water that is warmer, so when a body of water with colder, denser water is next to a body of water with warmer, less dense water, a density gradient is established. This gradient creates a difference in pressure between the two bodies of water, with the colder, denser water being at a higher pressure than the warmer, less dense water.
This difference in pressure creates a force that drives the movement of water from the denser, colder region to the less dense, warmer region. This movement of water is known as convection, and it can occur both vertically and horizontally in the ocean. Vertical convection occurs when differences in temperature cause water to rise or sink, while horizontal convection occurs when water moves laterally due to differences in temperature.
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missing options:
1. as water heats up, the atoms of water more faster.
2. warm water is pulled more by gravity than cold water.
3. warm and cold water mix and reach the same temperature.
4. bodies of water at different temperatures have different densities.
The compounds labeled benzophenone-3 (C14H12O3) and benzophenone-5 (C14H11NaO6S) are found in certain sunscreens. Would you expect a sunscreen made with benzophenone-3 or benzophenone-5 be more waterproof? Explain your choice.
A sunscreen made with Benzophenone-5 ([tex]C_1_4H_1_1NaO_6S[/tex]) would be expected to be more waterproof than benzophenone-3 ([tex]C_1_4H_1_2O_3[/tex]).
This is due to the presence of a sodium salt group (Na) and a sulfonic acid group ([tex]SO_3H[/tex] ) in benzophenone-5, which makes it more polar than benzophenone-3. Polar molecules interact more strongly with water molecules and are less likely to dissolve in nonpolar solvents such as oils.
Because sunscreen is designed to be water-resistant, the more polar benzophenone-5 should have stronger interactions with water and give more water resistance than benzophenone-3.
Moreover, the sulfonic acid group in benzophenone-5 may allow it to make stronger hydrogen bonds with water, increasing its water resistance even further.
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How many grams of air are in a 2.35 L balloon when its density is 1.4 g/L?
Answer:
3.29 grams
Explanation:
This is found by multiply 2.35 L by 1.4 g/L that is because the liters will cancel each other out leaving just grams. [tex]\frac{g}{L} * \frac{L}{1}[/tex]
A gas sample originally occupies 436 mL at 24 C. When the volume is expanded to 612 mL and the temperature is increased to 97 C, the pressure becomes 526 mm Hg. What was the original pressure?
Initially, there was a 266.8 mm Hg pressure.
solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas sample. The formula is:
(P1 × V1) ÷ (T1) = (P2 × V2) ÷ (T2)
where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
We are given that:
- V1 = 436 mL
- V2 = 612 mL
- T1 = 24 C + 273.15 = 297.15 K (convert from Celsius to Kelvin)
- T2 = 97 C + 273.15 = 370.15 K
- P2 = 526 mm Hg
We want to find P1, the original pressure.
Plugging in the values, we get:
(P1 × 436 mL) ÷ (297.15 K) = (526 mm Hg × 612 mL) ÷ (370.15 K)
Solving for P1, we get:
P1 = (526 mm Hg × 612 mL × 297.15 K) ÷ (436 mL × 370.15 K) = 266.8 mm Hg
Therefore, the original pressure was 266.8 mm Hg.
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Which of these are part of the
Earth's lithosphere?
O clouds
O glaciers
O mountains
O water vapor
Convert the following number
into correct scientific notation.
0.0602 x 10^25
[ ? ] × 10 [ ? ]
The number is converted to 60. 2 × 10²²
What are index forms?Index forms are simply described as mathematical forms that are used in the representation of numbers that are too small or too large in more convenient forms.
These index forms are also referred to as scientific notation or standard forms.
Some rules of index forms are;
Add the exponents when multiplying forms of the same basesSubtract the exponents when dividing forms of the same basesFrom the information given, we have that;
0. 0602 × 10 ²⁵
Subtract three from the exponent value and move three spaces right, we have;
60. 2 × 10²⁵⁻³
60. 2 × 10²²
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using the equation PCI5(g) PCI3(g) + CI2(g), if CI2 is added, what way will the euilibeium shift
When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.
Thus, The adjustments required to reach equilibrium might not be as obvious if we have a mixture of reactants and products that have not yet reached equilibrium.
In this situation, we can compare the Q and K values for the system to forecast changes.
By adding or withdrawing one or more of the reactants or products, an equilibrium chemical system can be momentarily moved out of equilibrium. After that, additional adjustments are made to the reactant and product concentrations in order to bring the system back to equilibrium.
Thus, When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.
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A solution contains 0.0400 M Ca2+ and 0.0990 M Ag+. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first?
Ca3(PO4)2
Ag3PO4
Na3PO4
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?
15.66% of the first cation is still in solution as the second cation is just beginning to precipitate.
What is phosphate used for?One of the three main nutrients that are most frequently used in fertilisers is phosphorous, which is obtained from processing phosphate rock (the other two are nitrogen and potassium).
You can also make phosphoric acid into phosphoric acids, which are utilised in everything from food and skincare to animal feed and electronics. Over the course of millions of years, organic matter accumulates to form the sedimentary rock known as phosphate.
When [tex]Na_{3}Po_{4}[/tex] added to the solution of Ca and Mg, [tex]Ca_{3}(Po_{4})_{2}[/tex] and [tex]Ag_{3}Po_{4}[/tex] are formed.
Ksp of [tex]Ca_{3}(Po_{4})_{2} = 2.07*10^{-33}[/tex]
Ksp of [tex]Ag_{3}Po_{4} = 0.09*10^{-17}[/tex]
Concentration of [tex][Ca^{2+}][/tex] = 0.040 M
Concentration of [tex][Ag^{+}][/tex] = 0.0990 M
[tex]Ag_{3} Po_{4} - > 3Ag^{+} + Po_{4}^{3-}[/tex]
Ksp = [tex][Ag^{+}]^{3} [Po4^{3-}][/tex]
[tex]0.09*10^{-17} = (0.099)^{3} [Po_{3-}][/tex]
[tex][Po_{3-}] = 9.16*10^{-14}M[/tex]
[tex]Ksp = [Ca^{2+}]^{3} [Po_{3-}][/tex]
[tex]2.07*10^{-33} = (0.040)^{3} [Po_{4}^{3-}]^{2}[/tex]
[tex][Po_{4}^{3-}] = 5.68*10^{-15} M[/tex]
[tex][Po_{4}^{3-}][/tex] is smaller in [tex]Ca_{3}(Po_{4})_{2}[/tex]
[tex]Ca_{3}(Po_{4})_{2}[/tex] will start precipitating first
[tex]Ksp = [Ca^{2+}]^{3} [Po_{4}^{3-}]^{2}[/tex]
[tex]2.07*10^{-33} = [Ca^{2+}]^{3} (9.16*10^{-14})^{2}[/tex]
[tex][Ca^{2+}] = 6.27*10^{-3} M[/tex]
[tex]\%\ of\ Ca^{2+}[/tex] remaining [tex]= 6.27*10^{-3}/0.040 * 100[/tex]
= 15.66 %
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3. In a lab, students mixed HCI acid with a Mg strip. The Mg started to bubble and dissolved within a few seconds. The rate at which the reaction occurs is determined by the A. number of effective collisions B. large AH C. the stabilization of the reactants D. mass of the products after the reaction
Answer:It might exposed
Explanation: or a spayed H2O might change because different water change over time
please help me pair pka values with displayed molecules
If we label the compounds ABCD from left to right;
A - 12.10
B - 15.90
C - 12.66
D - 12.35
What is the pKa?A molecule or compound's acidity is quantified by the pKa, which is the negative logarithm (base 10) of the acid dissociation constant (Ka). The lower the pKa value, the stronger the acid; it reflects a compound's propensity to give a proton (H+) in a solution.
The compound that has the highest number of attachment of the most electronegative elements would have the greatest pKa.
The justification of the answer above is that, seeing that the compound labelled B has three highly electronegative atoms hence it would have the most or the highest pKa of about 15.90 among the other compounds. The other compounds A, C and D have fewer electronegative atoms attached and thus a lower pKa as shown
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Pleas help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)
Moles of NA = Given Mass (g) ÷ Molecular Mass (g/mol)
= 27.5 ÷ 22.9897
= 1.196 mol
Moles of H2 Produced = Mol of NA × 1 mol H2 ÷ 2 Mol NA
= 1.196 × 1 ÷ 2
= 0.60 mol
Number of Molecules = Moles × Avogadro's Number
= 0.60 × 6.023 × 10²³ mol - 1
= 3.61 × 10²³
The Number of Molecules of Hydrogen Gas Produced When Added To Water Is 3.61 × 10²³
_________________________________
If the reaction A (aq) + B (aq) C(aq) has a Ka value equal to 4.26 x 10-6, what is the G value at 25 °C if the concentrations are as follows:
[A] = 1.50 M
[B] = 1.00 M
[C] = 5.00 x 10-5 M
The Gibbs free energy change for the given reaction at 25°C and the given concentrations is -25.5 kJ/mol
The Gibbs free energy change (∆G) of a reaction can be calculated using the equation:
∆G = -RT ln(K)
Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, and K is the equilibrium constant.
The equilibrium constant (K) can be calculated from the acid dissociation constant (Ka) as:
K = [C] ÷ ([A] × [B])
Substituting the given values, we get:
K = (5.00 x 10⁻⁵) ÷ (1.50 x 1.00) = 3.33 x 10⁻⁵
Therefore,
∆G = - (8.314 J/molK) × (298 K) × ln(3.33 x 10⁻⁵)
= 25.5 kJ/mol
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What is the difference between practical work inside a laboratory and outside a laboratory
Answer:
The main difference between practical work inside and outside a laboratory is that the practical work inside the lab includes good equipment and chemicals which are very advanced and the practical outside a laboratory is more about the safety of life.
Explanation:
Practicals are set up at stations with lab equipment and chemicals, where students can learn, and researchers can experiment and find different new things.
Thus, the practical work inside the lab includes lab equipment and chemicals, and the practical outside a laboratory is more about conserving nature.
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