The percentage of NaCl in the impure sample is 76.25%.
The given weight of an impure sample of table salt is 0.8421 g, and the weight of AgCl formed is 2.044 g. The percentage of NaCl in the impure sample can be determined as follows:
Calculations: Gram molecular weight of NaCl = 23 + 35.5 = 58.5 g
Number of moles of AgCl = weight of AgCl / gram molecular weight of AgCl= 2.044 / 143.5 = 0.0142 mol
The equation of the reaction between NaCl and AgNO3 is given as follows:
NaCl + AgNO3 → NaNO3 + AgCl
It can be seen from the above equation that 1 mole of NaCl gives 1 mole of AgCl.So, the number of moles of NaCl present in the sample is 0.0142 mol.
Percentage of NaCl in the sample = number of moles of NaCl present in the sample / number of moles of the impure sample × molar mass of NaCl × 100= (0.0142 / (0.8421 / 58.5)) × 100= 76.25%
Therefore, the percentage of NaCl in the impure sample is 76.25%.
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A sample of krypton gas in a container of volume 1.90 L exerts a pressure of 0.553 atm at 21°C. How many moles of gas are present?
Answer:
0.064 moles
Explanation:
First, we need to convert the temperature in Kelvin, which can be done by adding 273 to the temperature in Celsius. So the temperature in Kelvin is 294 K. We can now use the Ideal Gas Law, PV = nRT, to solve for the number of moles, n. Rearranging the equation gives us n = PV/RT. Plugging in our values gives us n = (0.553 atm)(1.90 L)/[(0.0821 L·atm/K·mol)(294 K)] = 0.064 mol. Therefore, there are 0.064 moles of krypton gas present in the container.
Which of the following compounds are most likely to be the same? Compound 1: 11 g of H and 89 g of 0. Compound 2: 20 g of H and 162 g of 0. Compound 3: 2 g of H and 33 g of 0. Compound 4: 8 g of H and 56 g of 0
Compounds 1 and Compound 2 are the same because they have the same empirical formula, which is H₂O.
What is the empirical formula of the compounds?All of the compounds have the same elements (H and O), but the ratios of the elements in each compound are different. To determine which compounds are the same, we need to calculate their empirical formulas.
The empirical formula of a compound gives the simplest whole-number ratio of the atoms in the compound.
To calculate the empirical formula, we divide the number of atoms of each element by its atomic weight and then divide by the smallest result.
Using this method, we get the following empirical formulas for each compound:
Compound 1: H₂O (divide 11 g of H by 1 g/mol and 89 g of O by 16 g/mol)
Compound 2: H₂O (divide 20 g of H by 1 g/mol and 162 g of O by 16 g/mol)
Compound 3: H₂O₄ (divide 2 g of H by 1 g/mol and 33 g of O by 16 g/mol)
Compound 4: H₂O₂ (divide 8 g of H by 1 g/mol and 56 g of O by 16 g/mol)
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In this assignment you will be asked to classify aqueous solutions of salts as to whether they are acidic, basic, or neutral. This is most easily done by first identifying how both the cation and anion affect the pH of the solution and then by combining the effects. After predicting the acid-base properties of these salts, you will then test your predictions in the laboratory. 1. State whether 0. 1 M solutions of each of the following salts are acidic, basic, or neutral. Explain your reasoning for each by writing ionic equations to describe the behavior of each salt in water: NACN, KNO3, NH4CI, NaHCO3, and Na3PO4,
NaCN: __________________
KNO3: _______________
NH4CI: ______________
NaHCO3: ______________
Na3PO4: ______________
0. 1 M solutions of each of the following salts are: NaCN will be basic, KNO₃ is Neutral, NH₄CI is Acidic, NaHCO₃ is Basic, and Na₃PO₄ will be Basic.
NaCN: Basic, When NaCN dissolves in water, it dissociates into Na+ and CN⁻. The CN⁻ ion can act as a weak base by reacting with water to form OH⁻ and HCN. It will be shown by the following equation:
CN⁻ + H₂O ↔ OH⁻ + HCN
Since the formation of OH⁻ ions leads to an increase in pH, NaCN solution is basic.
KNO₃: Neutral, When KNO₃ dissolves in water, it dissociates into K+ and NO₃⁻. Neither of these ions reacts with water to form H⁺ or OH⁻ ions. Therefore, the solution remains neutral.
NH₄CI: Acidic, When NH₄CI dissolves in water, it dissociates into NH₄⁺ and Cl-. The NH₄⁺ ion can act as a weak acid by reacting with water to form H₃O⁺ and NH₃. It will be shown by the following equation:
NH₄⁺ + H₂O ↔ H₃O⁺ + NH₃
Since the formation of H₃O⁺ ions leads to a decrease in pH, NH₄CI solution is acidic.
NaHCO₃: Basic, When NaHCO₃ dissolves in water, it dissociates into Na⁺ and HCO₃⁻. The HCO₃⁻ ion can act as a weak base by reacting with water to form H₂O and CO₃²⁻. It will be shown by the following equation:
HCO₃⁻ + H₂O ↔ H₂CO₃ + OH⁻ ↔ CO₃²⁻ + 2H₂O
Since the formation of OH⁻ ions leads to an increase in pH, NaHCO₃ solution is basic.
Na₃PO₄: Basic, When Na₃PO₄ dissolves in water, it dissociates into 3Na+ and PO₄³⁻. The PO₄⁻ ion can act as a weak base by reacting with water to form HPO₄⁻ and OH⁻. It will be shown by the following equation:
PO₄³⁻ + H₂O ↔ HPO₄²⁻ + OH⁻
Since the formation of OH⁻ ions leads to an increase in pH, Na₃PO₄ solution is basic.
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7. Explain how water in the Earth's polar regions can produce water vapour even
when the temperature is very low.
Where energy from the sun is enough to break the bonds between water molecules in the ice and convert them directly into water vapor without passing through the liquid phase.
What is Sublimation?
Sublimation is a physical process in which a solid substance is transformed directly into a gas without passing through the liquid phase. In other words, sublimation occurs when a solid substance changes its state directly to a vapor state, bypassing the liquid state.
Sublimation occurs when the vapor pressure of a solid exceeds the atmospheric pressure, causing the solid to transform into a gas without melting. This process requires energy to break the intermolecular bonds holding the solid together. This energy is typically supplied through heating or a decrease in pressure.
Water in Earth's polar regions can produce water vapor even when the temperature is very low due to a process called sublimation. Sublimation is a phase transition from a solid directly to a gas, without passing through the liquid phase. This process occurs when the vapor pressure of the solid (in this case, ice) exceeds the atmospheric pressure.
In the polar regions, the air is very cold and dry, with low atmospheric pressure. When the sun shines on the ice or snow, it provides energy that causes the surface layer of the ice to evaporate directly into water vapor without melting into liquid water.
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A certain amount of gas occupies 5.0 dm³ at 2 atm and 10⁰c. Calculate the number of moles present (R=0.082)
Answer:
n = 0.43 moles
Explanation:
from ideal gas equation;
PV = nRT
making n subject of the formula
n = PV/RT
but;
T = (10 + 273)k = 283k
P = 2 atm
V = 5dm³
R = 0.082
Therefore,
n = 2 x 5
0.082 x 283
= 0.43 moles
13. Solid sodium reacts violently with chlorine gas. The product formed in the reaction is sodium chloride, also
known as table salt. What type of reaction is this? Explain your answer.
The reaction of solid sodium and chlorine gas is a single displacement reaction resulting in sodium chloride.
Given that solid sodium reacts with chlorine gas to form sodium chloride also called as table salt.
The reaction is as follows: [tex]2Na + Cl2 -- > 2NaCl[/tex]
This is an example of a single displacement reaction, also known as a substitution reaction. In this type of reaction a more reactive element displaces a less reactive element from a compound.
Here, the sodium (a more reactive element) displaces the chlorine (a less reactive element) from the chloride compound, resulting in sodium chloride (table salt). The sodium atom donates one of its electrons to the chlorine molecule, forming a sodium cation (Na+) and a chloride anion (Cl-). They are then attracted to each other, forming the ionic compound sodium chloride.
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Calculate the amount of heat needed to convert 230.0g of ice at -10C to water at 0C
Answer:
It would take approximately 81.65 kJ (or 81,650 J) of heat to convert 230.0g of ice at -10C to water at 0C.
Explanation:
To calculate the amount of heat needed to convert ice at -10°C to water at 0°C, we need to consider two steps:
1. Heating the ice from -10°C to 0°C (heat required to raise the temperature of ice)
2. Melting the ice into water at 0°C (heat required to change the state of ice)
Let's first calculate the heat required for step 1:
Q1 = m × c × ΔT
where Q1 is the heat required, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
The specific heat capacity of ice is 2.09 J/g°C, and ΔT is (0°C - (-10°C)) = 10°C.
So, Q1 = 230.0 g × 2.09 J/g°C × 10°C = 4827 J
Now, let's calculate the heat required for step 2:
Q2 = m × Lf
where Q2 is the heat required, m is the mass of the ice, and Lf is the latent heat of fusion of ice.
The latent heat of fusion of ice is 334 J/g.
So, Q2 = 230.0 g × 334 J/g = 76820 J
Therefore, the total amount of heat needed to convert 230.0g of ice at -10C to water at 0C is:
Qtotal = Q1 + Q2 = 4827 J + 76820 J = 81647 J
Therefore, it would take approximately 81.65 kJ (or 81,650 J) of heat to convert 230.0g of ice at -10C to water at 0C.
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QUESTION IS NOT VISIBLE CLEARLY!!!!!!
Is it technically accurate to say a substance is insoluble? Why or why not? Provide an example of the Ksp value for a substance that is essentially insoluble
Yes, it is technically accurate to say a substance is insoluble if it has very low solubility in a particular solvent.
The Ksp value of AgCl is [tex]1.8 \times 10^{-10}[/tex].
The substance is insoluble means that the substance does not dissolve or dissolve only to a negligible extent in the solvent.
For example, silver chloride (AgCl) is considered an insoluble substance because it has very low solubility in water. The solubility product constant (Ksp) for AgCl in water at room temperature is approximately [tex]1.8 \times 10^{-10}[/tex]
This means that the concentration of silver ions ([tex]Ag^+[/tex]) and chloride ions ([tex]Cl^-[/tex]) in a saturated solution of AgCl is very low, indicating that only a small amount of AgCl has dissolved in the water.
We can say that AgCl is essentially insoluble in water, and any small amount of AgCl that does dissolve in water will immediately re-precipitate out of the solution once the solution becomes saturated with respect to AgCl.
Therefore, we can say that AgCl is essentially insoluble in water, and any small amount of AgCl that does dissolve in water will immediately re-precipitate out of the solution once the solution becomes saturated with respect to AgCl.
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Write 36.82 correct to the nearest whole number.
Answer:
36.82 is 37 when you round it
Explanation:
:)
mass fraction of water in a solution is 0.8. what is the volume of water containing 150g of sucrose
Answer:
The volume of water containing 150g of sucrose is 600 mL.
Step by step explanation:
To solve this problem, we need to first determine the total mass of the solution. We know that the mass fraction of water is 0.8, which means that water makes up 80% of the solution.
Therefore, the mass fraction of the other component (sucrose) is 0.2, or 20%.
Let's assume we have a total mass of 1 kg (1000 grams) of solution, then the mass of water in the solution would be:
Mass of water = 0.8 x 1000 grams = 800 gramsSince the mass of sucrose is 20% of the total mass,
we can calculate it as:
Mass of sucrose = 0.2 x 1000 grams = 200 grams
Now we can use the density of water to calculate the volume of water that contains 800 grams.
The density of water is approximately 1 gram per milliliter (g/mL).
Therefore, the volume of water that contains 800 grams is:
Volume of water = 800 grams / 1 g/mL = 800 mL
Finally, we can calculate the volume of water that contains 150 grams of sucrose by using the mass ratio of sucrose to water, which is:
150 grams of sucrose / 200 grams of sucrose = 0.75
This means that the volume of water containing 150 grams of sucrose is 0.75 times the volume of water in the entire solution:
Volume of water containing 150g of sucrose = 0.75 x 800 mL = 600 mL
Therefore, the volume of water containing 150g of sucrose is 600 mL.
How does the amount of salt affect the density of the water?
Answer:
the water sample with higher salinity will have greater mass, and it will therefore be more dense.
Explanation:
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the volume of a certain gas was found to be 1600cm3 when the pressure was 800mmHg. At what pressure the volume will be decreased to 80% of the original volume.
The pressure will be 1000 mmHg when the volume is decreased to 80% of the original volume.
What is Boyle's law ?Boyle's law, which states that at a constant temperature, the pressure and volume of a gas are inversely proportional.
We can use Boyle's law. This means that:
P 1V1 = P2V2
Where
P1 and V1 are the initial pressure and volume P2 and V2 are the final pressure and volumeIn this case, we know that the initial pressure (P1) is 800 mmHg and the initial volume (V1) is 1600 cm3. We want to find the final pressure (P2) when the volume (V2) is decreased to 80% of the initial volume:
V2 = 0.8V1 = 0.8 x 1600 cm3 = 1280 cm3
Substituting these values into Boyle's law, we get:
P1V1 = P2V2
800 mmHg x 1600 cm3 = P2 x 1280 cm3
Solving for P2, we get:
P2 = (800 mmHg x 1600 cm3) / 1280 cm3
P2 = 1000 mmHg
Therefore, the pressure will be 1000 mmHg when the volume is decreased to 80% of the original volume.
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If 5. 00L of nitrogen reacts completely with hydrogen at a pressure of 3. 00atm and a temperature of 298k, how much ammonia, in gram, is produced?
If 5.00L of nitrogen completely reacts with hydrogen at a pressure of 3.00atm and a temperature of 298k,approximately 20.67 grams of ammonia will be produced.
The balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia is:
N2 + 3H2 → 2NH3
We can use the ideal gas law to calculate the number of moles of nitrogen:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L atm mol K), and T is the temperature in Kelvin.
Rearranging the equation, we get:
n = PV/RT
Substituting the given values, we get:
n = (3.00 atm) (5.00 L) / (0.0821 L atm mol K) (298 K)
n = 0.607 mol
n(NH3) = 2 × n(N2) = 2 × 0.607 mol = 1.214 mol
Mass(NH3) = n(NH3) × M(NH3) = 1.214 mol × 17.03 g/mol ≈ 20.67 g
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How much KNO3 must be
dissolved in 50 g of water to
yield a saturated solution at
50°C?
Answer:
The solubility of KNO3 in water increases with temperature. At 50°C, the solubility of KNO3 in water is 34 g/100 g water. This means that at 50°C, 100 g of water can dissolve up to 34 g of KNO3.
To find out how much KNO3 must be dissolved in 50 g of water to yield a saturated solution at 50°C, we can set up a proportion:
34 g KNO3/100 g water = x g KNO3/50 g water
where x is the amount of KNO3 that must be dissolved in 50 g of water.
Solving for x, we get:
x = (34 g/100 g) x 50 g
x = 17 g
Therefore, 17 g of KNO3 must be dissolved in 50 g of water to yield a saturated solution at 50°C.
Explanation:
Answer:
Explanation:
This is the simple answer of this question with respect to 60 degree temperature but question requirement is 50 so we will do it according to 50 degree temperature
the concentration of one sample of dissolved brass solution was determined to contain 0.0205 m cu2 . if the volume of the sample is 50.0 ml, how many moles cu2 are in the sample?
There are 0.001025 moles of Cu2+ in the sample. Since the concentration of the dissolved brass solution is given as 0.0205 M (molar concentration) for Cu2+.
What are the brass's solvent and solution?Brass is a solid mixture (alloy) made up of 30% zinc and 70% copper. The solvent is the material in greater quantity, and the solute is the substance in smaller quantity. Zinc serves as the solute while copper serves as the solvent in brass.
We can use the following formula to calculate the number of moles of Cu2+ in the sample:
moles Cu2+ = concentration (M) x volume (L)
However, we need to convert the volume from milliliters (mL) to liters (L) in order to use this formula. We can do this by dividing the volume by 1000:
50.0 mL = 50.0 / 1000 L = 0.0500 L
Now we can plug in the values and solve for the number of moles:
moles Cu2+ = 0.0205 M x 0.0500 L
moles Cu2+ = 0.001025 mol
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what happens if more sodium chlorate is added increase or decrease balanced equation sodium chlorate decomposes to produce sodium chloride plus oxygen gas
When more sodium chlorate is added, it will increase the amount of products formed in the balanced equation: Sodium chlorate decomposes to produce sodium chloride plus oxygen gas.
The balanced equation for this reaction is: 2NaClO3 → 2NaCl + 3O2.In a chemical reaction, the reactants (sodium chlorate in this case) are transformed into products (sodium chloride and oxygen gas). According to the law of conservation of mass, the total mass of reactants must equal the total mass of products.
Therefore, the coefficients in a balanced chemical equation must always be proportional and consistent with the number of atoms/molecules in the reactants and products. Since adding more reactant will result in the formation of more products, the coefficients in the balanced equation will need to be adjusted to reflect this increase.
For example, if more sodium chlorate is added, the coefficients for sodium chloride and oxygen gas will increase accordingly to maintain a balanced equation. The equation will look something like this: 2NaClO3 + x → 2NaCl + 3O2 where x represents the additional amount of sodium chlorate added.
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the table shows a list of compounds that are gases at stp identify the type of bonding (ionic, covalent, or metallic) for each compound listed
hydrogen sulfide (H2S)
Ammonia (NH3)
Methane (CH4)
Nitrous Oxide (N2O)
Answer:
hydrogen sulfide (H2S): covalent
Ammonia (NH3): covalent
Methane (CH4): covalent
Nitrous Oxide (N2O): covalent
Explanation:
o Hydrogen sulfide (H2S is a covalent compound that is composed out of 2 hydrogen atoms.
o Because one nonmetal combines with another nonmetal to form a covalent compound, ammonia (NH3) is a covalent compound.
o CH4 (Methane) is a covalent (nonpolar covalent) compound because when two nonmetals combine, they form a covalent compound.
o Nitrous Oxide (N2O) is a covalent bond. Since the N-N-O bond angle is 1800, the central N makes one covalent bond with N and O.
Answer:
All of the compounds listed above have covalent bonding.
according to science direct,
A covalent bond consists of the mutual sharing of one or more pairs of electrons between two atoms. These electrons are simultaneously attracted by the two atomic nuclei.
all of the compounds above share 1 or more pairs of electrons.
44 grams of carbon dioxide, CO2 is dissolved in 5000 ml of solution
The concentration of the carbon dioxide solution is 8.8 g/L.
What is Solution?
A solution is a homogeneous mixture of two or more substances. In a solution, the particles of the solute (the substance that is dissolved) are uniformly distributed throughout the solvent (the substance in which the solute is dissolved) on a molecular level. Solutions can be in any physical state, i.e., solid, liquid, or gas. In a solution, the solute particles are too small to be seen, and the solution appears clear and transparent.
To determine the concentration of the carbon dioxide solution, we need to know the units of the given mass (44 grams). Let's assume it is in grams per liter (g/L).
First, we need to convert the volume of the solution from milliliters to liters:
5000 mL = 5 L
Next, we can use the formula for concentration:
concentration (in g/L) = mass (in g) / volume (in L)
Plugging in the values we get:
concentration = 44 g / 5 L
Simplifying, we get:
concentration = 8.8 g/L
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In your double replacement reaction, the more reactive metal "pushed" the other one out of its place. Without using the Internet, where could you look to determine which metal was more reactive?
Answer:
Explanation:
To determine which metal is more reactive, you could refer to the reactivity series of metals. The reactivity series is a list of metals arranged in order of their reactivity with the most reactive metal at the top of the list and the least reactive metal at the bottom.
If you have a copy of a chemistry textbook, you could refer to the chapter on the reactivity of metals or look up the reactivity series in the index. Alternatively, you could consult a periodic table that includes the relative reactivity of the elements. The reactivity series is also often included in reference materials such as handbooks, tables, and charts of chemical data.
the rate of a standard reaction is 0.00543 m/s at 40 oc. what will the rate be if the temperature is doubled?
The rate of the reaction will also double and become 0.01068 m/s.
The Arrhenius equation can be used to determine the effect of temperature on reaction rates. A reaction's rate constant is related to temperature by the Arrhenius equation, which is as follows:
[tex]k = Ae^(^-^E^a^/^R^T^)[/tex]
where: k = rate constant
A = frequency factor
e = base of natural logarithms (2.71828...)E
a = activation energy
R = gas constant (8.314 J/(mol K))
T = temperature (in Kelvin)
Since k is directly proportional to the rate of a reaction, we can use the equation to determine the effect of temperature on the rate of a reaction. Doubling the temperature will result in the following changes:
If the temperature is doubled, [tex]T_2 = 2T_1[/tex]
The equation becomes:
[tex]k_2 = Ae^(^-^E^a^/^R^T^2)[/tex]
We can now substitute 2T1 for T2:
[tex]k_2 = Ae^(^-^E^A^/^R^(^2^T1^)[/tex]
Simplify:k2 = [tex]Ae^(^-^E^a^/^R^T^1) * e^(^-^E^a^/^R^T^1^)[/tex]
[tex]k_2 = k_1 * e^(^-^E^a^/^R^T^1^) * e^(^-^E^a^/^R^T^1^)[/tex]
[tex]k_2 = k_ 1* e^(^-^2^E^a^/^R^T^1^)[/tex] So, the rate constant doubles if the temperature is doubled since [tex]e^(^-^2^E^a^/^R^T^1^)[/tex] is positive.
Therefore, the rate of a standard reaction will be 0.00543 m/s x 2 = 0.01086 m/s if the temperature is doubled.
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22. 0 mL of stock solution is used
to produce a. 458 M solution after
dilution with 50. 0 mL of water. What
is the molarity of the stock solution?
The molarity of the stock solution is 1.05 M.
We can use the equation for dilution to solve this problem:
M₁V₁ = M₂V₂
where M₁ is the initial molarity (of the stock solution), V₁ is the initial volume (in mL) of the stock solution, M₂ is the final molarity (after dilution), and V₂ is final volume (in mL) of diluted solution.
In this case, we know that V₁ = 22.0 mL, V₂ = 50.0 mL, and M₂ = 0.458 M. We want to solve for M₁.
Plugging in the values, we get:
M₁(22.0 mL) = (0.458 M)(50.0 mL)
Simplifying and solving for M₁, we get:
M₁ = (0.458 M)(50.0 mL) / (22.0 mL)
M₁ = 1.05 M
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What happens to the mass of a metal when it reacts with oxygen?
Answer:
it gains mass
Explanation:
metals have incorporated oxygen atoms
what is metallic lattice?
30 POINTS PLEASE HELP
What volume of a 0.500 M sodium chloride solution is required to make 500. mL of a 0.100 M sodium chloride solution?
A)1 L
B)100 L
C)0.100 ml
D)0.100 L
Answer:
D- 0.100L
Explanation:
you are instructed to reflux the reaction for 1 hour. if the instructions simply told you to reflux until the reaction was complete, how would you know when this was?
If the instructions simply told you to reflux until the reaction was complete, you should know when the reaction is complete by observing the reaction progress.
To determine whether or not the reaction is finished, you may rely on various physical and chemical characteristics, such as the reaction colour, temperature, pressure, pH, and solvent composition.
The reaction completion point is the moment at which no further changes in the reaction mixture can be observed or when the reaction mixture reaches a state of equilibrium.
Some typical methods of observing the completion of a reaction include the following: Monitoring the pH by using a pH meter checking the temperature of the solution ensuring the pressure in the reaction vessel to monitor the gas evolution tracking the colour of the reaction mixture.
To summarize when you're told to reflux until the reaction is complete, the only way to know when that happens is to look for observable changes. You should monitor the progress of the reaction by observing the changes in physical and chemical properties.
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Which factor least influences the rate of photosynthesis
Answer:
light intensity
Explanation:
Convert 50 g of chlorine gas to molecules
1.410317885651
Below me are the examples:
1 grams Chlorine to mol = 0.02821 mol
10 grams Chlorine to mol = 0.28206 mol
20 grams Chlorine to mol = 0.56413 mol
30 grams Chlorine to mol = 0.84619 mol
40 grams Chlorine to mol = 1.12825 mol
50 grams Chlorine to mol = 1.41032 mol
100 grams Chlorine to mol = 2.82064 mol
200 grams Chlorine to mol = 5.64127 mol
What is the number of ions in 0. 20 mol of (NH4)3PO4?
The number of ion in 0.2 mole of (NH₄)₃PO₄ is found to be 2.4 x 10²³.
The number of ions and the number of moles of a compound are related to each other as,
Moles = Number of ions/Avogadro number = molar mass/given mass
Avogadro number = 6.022 x 10²³
Here, (NH₄)₃PO₄ will give two ions after dissociation will give two ions, so, the number of moles will be equal to two times of the number of particles of the compound. So, the total moles of ions will be 0.4 moles.
Putting all the values is the formula,
6.022 x 10²³ x 0.4 = Number of ions
So, finally after solving, we get, 2.4 x 10²³ as the number of ions in the compound of (NH₄)₃PO₄.
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Example: PuC2Prunium corninePrunium cornide
Name these Ionic Compounds using the “Periodic Table of Food”:
2. BPo
3. Bl2Tu
4. Cr2Sn
5. LiSr2
6. Or3Ba2
The name of the ionic compounds include:
2. BPo is Berry Polonium oxide
3. Bl₂Tu is Blueberry Tungsten dichloride
4. Cr₂Sn is Cranberry Tin oxide
5. LiSr₂ is Lime Strontium dichromate
6. Or₃Ba₂ is Orange Barium trioxide
What is Ionic Compounds?Ionic compounds are chemical compounds formed by the electrostatic attraction between positively and negatively charged ions. They are composed of positively charged ions, called cations, and negatively charged ions, called anions. Ionic compounds are typically formed by the reaction of a metal with a nonmetal, or of a metal and a polyatomic ion.
In an ionic compound, the number of positive charges must equal the number of negative charges so that the compound has a neutral overall charge. Ionic compounds tend to have high melting and boiling points, as well as good electrical conductivity in their molten or dissolved states, but are typically brittle solids at room temperature.
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