How many moles of AgNO3 are in 338 grams of AgNO?

Answer

A

Explanation

letter a kase tama

Answer:

Fluroine

Explanation:

Answer:

Fluorine

Explanation:

Nitrogen atomic radius: 155 pm

Oxygen atomic radius: 152 pm

Fluorine atomic radius: 147 pm

Carbon atomic radius: 170 pm

Therefore, fluorine has the smallest atomic radius.

Answer:

It's very similar to it

Explanation:

Because the atomic mass in this situation is an average so it represents all isotopes in this situation due to their abundance

Atomic mass is an average mass of an element whereas the mass number is a mass of a single atom of an element.

Atomic mass

Mass number

So, from this, we can conclude that atomic mass is an average mass of an element whereas the mass number is a mass of a single atom of an element.

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The calculated molarity is 2.5 M. Molarity (M), which is the quantity of solute in moles divided by the volume of solution, is the most widely used unit to express solution concentration.

The amount of a substance in a specific volume of solution is known as its molarity (M). The number of moles of a solute per liter of a solution is known as molarity. The molar concentration of a solution is another name for molarity.

Molality is defined as the product of the number of moles of solute and the mass of the solvent in kilograms (kg).

NaOH has a mass of 10g, a molar mass of 40 g/mol, and a mass of 100g, or 0.1 kg, for water.

Molality is therefore 10/400.1 = 2.5 m.

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A. because he might be in trouble in a play

To determine which of the statements are true after the bacteria have had a chance to do their work, we need to examine the equation for denitrification and consider the initial concentrations of O2, NO3-, and CH2O in the wastewater.

From the denitrification equation, we see that 4 moles of NO3- react with 5 moles of CH2O to produce 2 moles of N2, 5 moles of CO2, and 7 moles of H2O. The H+ ions in the equation simply balance the charges, so we can ignore them for this analysis.

Let's first calculate the moles of NO3- and CH2O in the wastewater:

Moles of NO3- = (1 x 10^-3 M) x (1 L) = 1 x 10^-3 mol

Moles of CH2O = (1.00 x 10^-2 M) x (1 L) = 1.00 x 10^-1 mol

Now let's consider the O2 concentration. We don't have an equation for aerobic respiration, but we know that it consumes O2 and produces CO2 and H2O. If the bacteria in the container are consuming organic matter, they are likely using aerobic respiration initially until the O2 is depleted. So, if there is any O2 remaining after the bacteria have had a chance to do their work, it would suggest that aerobic respiration was the dominant process.

The initial O2 concentration is 8 mg/L, but we need to convert that to moles/L to compare it to the moles of NO3- and CH2O:

Molecular weight of O2 = 32 g/mol

Moles of O2 = (8 mg/L) / (32 g/mol) / (1000 mL/L) = 2.5 x 10^-4 mol/L

Comparing the moles of O2 to the moles of NO3- and CH2O, we see that there are far fewer moles of O2 than either NO3- or CH2O. Therefore, we can expect that aerobic respiration will have consumed most of the O2, and denitrification will be the dominant process for consuming the organic matter.

To confirm this, we can compare the moles of NO3- and CH2O to the stoichiometry of the denitrification equation:

Moles of NO3- / 4 = 2.5 x 10^-4 mol/L / 4 = 6.25 x 10^-5 mol/L

Moles of CH2O / 5 = 1.00 x 10^-1 mol/L / 5 = 2.00 x 10^-2 mol/L

The moles of NO3- are much smaller than the moles of CH2O, so denitrification will consume most of the organic matter. The products of denitrification are N2, CO2, and H2O, so we can expect that the concentrations of NO3- and CH2O will decrease, the concentration of O2 will decrease (due to aerobic respiration), and the composition of the water will change. Specifically, we can expect that:

(a) No CH2O will remain - This statement is not true. There will likely be some CH2O remaining, since the equation for denitrification consumes only a fraction of the initial concentration of CH2O.

(b) Some O2 will remain - This statement is not true. The initial concentration of O2 is much smaller than the initial concentrations of NO3- and CH2O, so it will be consumed quickly by aerobic respiration

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Answer:

Knowing chemistry can help you make decisions

Answer:

1 ) chemistry plays a role in the manufacturing of medicines

2 ) chemistry plays a role in the production of cosmetics

3 ) chemistry plays a role in the production of soaps and detergents

4 ) chemistry is used to manufacture construction materials

5 ) chemistry allows us to use fuels

6 ) chemistry is important in agriculture ( providing fertilizers ... )

7 ) chemistry is important in textiles

The retention factor for the yellow spot is 0.2586.

Rf = Distance traveled with the aid of the compound / Distance traveled by the solvent the front

Rf = 1.5 cm / 5.8 cm = 0.2586

Retention factor (Rf) is a term used in chromatography, a laboratory technique used to separate and analyze mixtures of compounds. Rf is a dimensionless quantity that describes the migration of a particular compound relative to the solvent front in a chromatographic system.

In a typical chromatographic separation, the sample mixture is placed on a stationary phase, which is usually a solid or liquid that is immobilized on a solid support. The stationary phase is then contacted with a mobile phase, which is a liquid or a gas that flows over the stationary phase, carrying the sample compounds with it. The Rf value for a given compound is determined by dividing the distance the compound travels up the stationary phase by the distance the solvent front travels up the same phase.

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Six more hydrogen atoms does a cyclohexane molecule have than a benzene molecule.

Benzene (C₆H₆) is a cyclic hydrocarbon consisting of a hexagonal ring of carbon atoms with alternating single and double bonds. Each carbon atom in the benzene ring is bonded to one hydrogen atom. Therefore, a benzene molecule contains six carbon atoms and six hydrogen atoms.

Therefore, benzene contains six carbon atoms and six hydrogen atoms.

Cyclohexane (C₆H₁₂), on the other hand, is a saturated hydrocarbon known as a cycloalkane. It also consists of a hexagonal ring of carbon atoms, but unlike benzene, all the carbon-carbon bonds in cyclohexane are single bonds. This means that each carbon atom in the cyclohexane ring is bonded to two additional hydrogen atoms compared to benzene

Therefore, cyclohexane contains six carbon atoms and 12 hydrogen atoms.

The difference in the number of hydrogen atoms between a cyclohexane molecule and a benzene molecule is 12 - 6 = 6. Hence, a cyclohexane molecule has six more hydrogen atoms than a benzene molecule.

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The likelihood of side reactions and the formation of byproducts during the nitration of methyl benzoate increases at higher temperatures due to the increased reactivity and concentration of the reactive species, and the decreased stability of the nitronium ion.

In the nitration of methyl benzoate reaction, the reaction is typically carried out by adding a mixture of concentrated nitric acid and sulfuric acid to methyl benzoate. This reaction is exothermic and the heat generated from the reaction can cause the reaction mixture to heat up.

At higher temperatures, there is an increased likelihood of side reactions occurring. One common side reaction that can occur during the nitration of methyl benzoate is the formation of byproducts such as dinitromethyl benzoate and trinitromethyl benzoate. These byproducts are formed when excess nitric acid reacts with methyl benzoate, rather than just a single nitro group being added to the benzene ring.

The formation of these byproducts is favored at higher temperatures because the reaction rate of the nitration reaction increases with temperature, and the concentration of the reactive species is higher at higher temperatures. This means that there is a greater likelihood of excess nitric acid being present and reacting with the methyl benzoate at higher temperatures, which can result in the formation of the unwanted byproducts.

Furthermore, the stability of the nitronium ion, which is the key reactive species in the nitration of methyl benzoate, decreases at higher temperatures. This can lead to the formation of other unwanted side products, such as benzene, due to the decomposition of the nitronium ion.

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Polywool fibre is synthetic.

⇒b. Synthetic

Hope It Helps You ✌️

Answer:

\(\large{\underbrace{\underline{\fcolorbox{White}{pink}{\bf{b.synthetic }}}}}

\)

it’s C

i hope i help you but if not DO NOT FORGET STAY SAFE AT HOME

Answer:

D.

Explanation:

The electrons farthest from the nucleus of the atom​

The percent yield of the reaction is 84.62%.

To calculate the percent yield of a reaction, you need to know the theoretical yield and the actual yield of the product.

In this case, the balanced equation for the reaction between CH4 and steam (H2O) is:

CH4 + 2H2O → CO2 + 4H2

From the equation, we can see that for every mole of CH4 reacted, we should get 4 moles of H2 produced.

To determine the theoretical yield of H2, we need to convert the given mass of CH4 to moles and then use the mole ratio from the balanced equation to calculate the expected amount of H2 produced.

Molar mass of CH4 = 16 g/mol

Number of moles of CH4 = 61,500 g / 16 g/mol = 3843.75 mol

From the balanced equation, 1 mole of CH4 produces 4 moles of H2.

So, the expected moles of H2 = 3843.75 mol x 4 = 15375 mol

The actual yield of H2 is given as 13.0 kg = 13,000 g.

Now, we can calculate the percent yield using the following formula:

Percent yield = (Actual yield / Theoretical yield) x 100%

Plugging in the values we obtained above, we get:

Percent yield = (13,000 g / 15375 mol) x 100%

= 84.62%

Therefore, the percent yield of the reaction is 84.62%.

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To complete and balance the molecular equation for the reaction of aqueous chromium(II) bromide (CrBr₂) and aqueous sodium carbonate (Na₂CO₃). Here is the step-by-step explanation:

1. Write the unbalanced molecular equation, including the physical states:
CrBr₂(aq) + Na₂CO₃(aq) → ?

2. Determine the products of the reaction. Chromium(II) bromide will react with sodium carbonate to form chromium(II) carbonate (CrCO₃) and sodium bromide (NaBr):
CrBr₂(aq) + Na₂CO₃(aq) → CrCO₃(s) + NaBr(aq)

3. Balance the equation:
CrBr₂(aq) + Na₂CO₃(aq) → CrCO₃(s) + 2 NaBr(aq)

So, the balanced molecular equation for the reaction of aqueous chromium(II) bromide and aqueous sodium carbonate is:
CrBr₂(aq) + Na₂CO₃(aq) → CrCO₃(s) + 2 NaBr(aq)

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The approximate resistance value at 45 degrees Celsius is around 5.8 ohms.

To calculate the value of the resistance temperature coefficient at 45 degrees Celsius using linear approximation, we can use the formula:

R(T) = R0 + α(T - T0),

where R(T) is the resistance at temperature T, R0 is the resistance at a reference temperature T0, α is the resistance temperature coefficient, and (T - T0) is the temperature difference.

Given that the resistance at 25 degrees Celsius is 5 ohms (R0 = 5) and the resistance at 50 degrees Celsius is 6 ohms (R(T) = 6), we can calculate the value of α.

6 = 5 + α(50 - 25),

Simplifying the equation:

1 = 25α,

Therefore, α = 1/25 = 0.04 ohm/degree Celsius.

Using the linear approximation, we can approximate the value of the resistance at 45 degrees Celsius:

R(45) = 5 + 0.04(45 - 25) = 5 + 0.04(20) = 5 + 0.8 = 5.8 ohms.

Therefore, the value of the resistance at 45 degrees Celsius is approximately 5.8 ohms.

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Answer: try 42.8

Explanation: you might have to round to 42.8 because of the significant  figures

Hello! HNO2 is not a strong acid so therefore your answer is A.

A rule of thumb, the rest are strong acids, so when you ever come across a similar question you will be able to rule out the wrong answers quicker :).

Strong acid list:

H2SO4

HNO3

HCLO4

HCLO3

HCL

HBr

HI

hope this helps!

17.85 mL of 3.0 M stock solution of nitric acid is needed to make a 150 mL 0.357 M nitric acid.

The dilution of a stock solution can be solved by using the formula below, which indicates that the concentration of the solution changes but the number of moles of the solute remains the same.

(concentration of stock solution)(volume of stock solution) = (concentration of diluted solution)(volume of diluted solution)

Given:

(3.0 M)(volume of stock solution)=(0.357 M)(150 mL)

volume of stock solution = [(0.357 M)(150 mL)] / (3.0 M)

volume of stock solution = 17.85 mL

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Answer:  330.7 jule energy

A direct titration of the acetylsalicylic acid in the tablet with sodium hydroxide is not performed because of C. Acetylsalicylic acid is unstable in the presence of sodium hydroxide and will undergo decomposition releasing toxic benzene - to which we do not want our students exposed.

Acetylsalicylic acid is unstable in the presence of sodium hydroxide and will undergo decomposition releasing toxic benzene - to which we do not want our students exposed.

This means that the reaction between acetylsalicylic acid and sodium hydroxide is not a simple acid-base reaction and will result in the formation of other products, including toxic benzene. Therefore, it is not safe to perform this titration in a classroom or laboratory setting.

It is important to note that the other options provided are not correct. Acetylsalicylic acid is not a diprotic acid (option A), it is a weak acid but it does react with sodium hydroxide (option B), and it does not undergo more than one reaction with sodium hydroxide (option D).

In conclusion, a direct titration of the acetylsalicylic acid in the tablet with sodium hydroxide is not performed because the reaction is not a simple acid-base reaction and will result in the formation of toxic benzene.

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Based on the information provided, the principle of the conservation of mass did apply in this example.

The principle of the conservation of mass states that mass is neither created nor destroyed in a chemical reaction. In other words, the total mass of the reactants should be equal to the total mass of the products.

In the given scenario, Nicole measured 25 g of sodium carbonate and 10 mL of vinegar, which can be considered the reactants. The total mass of the reactants and the beaker was determined to be 100 g. After mixing the reactants, bubbling and a white residue were observed, and the total mass became 98 g.

To analyze the conservation of mass, we need to consider the mass of the products formed. The bubbling and white residue suggest a chemical reaction occurred, likely resulting in the formation of a gas and a solid product. Although the exact reaction and products are not specified, it is evident that some change took place.

The total mass decreasing from 100 g to 98 g indicates that the mass of the products is less than the mass of the reactants and the beaker. This might be due to the formation of a gas that escaped from the reaction mixture.

While the total mass decreased, it is important to note that mass was not created or destroyed. The lost mass in the form of the escaping gas can be accounted for if it is considered separately.

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The percent yield of C₆H₁₂O₆ is 100.13%.

To calculate the percent yield of C₆H₁₂O₆, we need to compare the actual yield (7.5 g) of C₆H₁₂O₆ with the theoretical yield of C₆H₁₂O₆ that can be calculated from the given reactants (12 g CO₂ and 8.0 g H₂O).

Firstly, we calculate the molar masses of the reactants and the product.

The molar mass of CO₂ = 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol

The molar mass of H₂O = 2 × 1.01 g/mol + 16.00 g/mol = 18.02 g/mol

The molar mass of C6H12O6 = 6 × 12.01 g/mol + 12 × 1.01 g/mol + 6 × 16.00 g/mol = 180.18 g/mol

Now, Convert the given masses of reactants and product to moles.

Moles of CO₂ = 12 g / 44.01 g/mol = 0.2726 mol

Moles of H₂O = 8.0 g / 18.02 g/mol = 0.4439 mol

Moles of C₆H₁₂O₆ (product) = 7.5 g / 180.18 g/mol

= 0.0416 mol

Now, we can determine the limiting reactant.

Since the reaction ratio between CO₂ and C₆H₁₂O₆ is 6:1, and the moles of CO₂ (0.2726 mol) are greater than 6 times the moles of C₆H₁₂O₆(0.0416 mol), CO₂ is the limiting reactant.

Now, we can calculate the theoretical yield of C₆H₁₂O₆

Theoretical yield of C₆H₁₂O₆ = moles of limiting reactant (CO₂) × molar mass of C₆H₁₂O₆

= 0.0416 mol × 180.18 g/mol = 7.49 g

Finally, we can calculate the percent yield.

Percent yield = (actual yield / theoretical yield) × 100

= (7.5 g / 7.49 g) × 100

= 100.13%

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Approximately 24.55 cm^3 of oxalic acid must be added to sufficient water to give a 1.500 liter solution with a formal concentration of 0.300 F.

To find the volume of oxalic acid needed to make a 1.500 liter solution with a formal concentration of 0.300 F, we need to use the equation:

Formal concentration (F) = (moles of solute) / (volume of solution in liters)

First, we need to calculate the moles of oxalic acid required. The formal concentration (F) is given as 0.300, so:

0.300 = (moles of oxalic acid) / 1.500

Rearranging the equation, we find:

moles of oxalic acid = 0.300 * 1.500

moles of oxalic acid = 0.450

Next, we can calculate the mass of oxalic acid needed using its molecular mass:

mass of oxalic acid = moles of oxalic acid * molecular mass

mass of oxalic acid = 0.450 * 90.03

mass of oxalic acid = 40.5145 g

Finally, we can calculate the volume of oxalic acid needed using its density:

volume of oxalic acid = mass of oxalic acid / density

volume of oxalic acid = 40.5145 g / 1.65 g/cm^3

volume of oxalic acid = 24.55 cm^3

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Answer:

Avogadro's law.

Explanation:According to Avogadro's law,at constant temperature and pressure,the volume of gas is directly proportional to the number of moles of the gas.

The new boiling point of the NaCl solution is 102.91°C when 0.125 kg of NaCl is added to 750 g of water, assuming the boiling point elevation constant (Kb) of water is 0.51°C/m.

To find the new boiling point of the solution, we can use the formula:

ΔTb = Kb × m

where ΔTb is the change in boiling point, Kb is the boiling point elevation constant (0.51°C/m for water), and m is the molality of the solute.

First, we need to calculate the molality of the NaCl solution:

m = (moles of solute) / (mass of solvent in kg)

The molar mass of NaCl is 58.44 g/mol, so 0.125 kg of NaCl is:

moles of NaCl = (0.125 kg) / (58.44 g/mol) = 2.14 mol

The mass of the solvent (water) is 750 g, which is 0.75 kg.

So, the molality of the NaCl solution is:

m = (2.14 mol) / (0.75 kg) = 2.85 mol/kg

Now we can use the formula to calculate the change in boiling point:

ΔTb = Kb × m = (0.51°C/m) × (2.85 mol/kg) = 1.454°C

Finally, we can find the new boiling point of the solution by adding the change in boiling point to the normal boiling point of water (100°C):

New boiling point = 100°C + 1.454°C = 101.454°C

Therefore, the correct answer is c. 102.91°C.

The new boiling point of the NaCl solution is 102.91°C when 0.125 kg of NaCl is added to 750 g of water, assuming the boiling point elevation constant (Kb) of water is 0.51°C/m.

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Answer:Atoms seek to be stable; so, to get to a more stable state, the atom expels energy from the nucleus in the form of a particle or ray. This process is known as radioactivity, the unstable atom is said to be a radioactive atom, and the energy that's released is radiation

Explanation: so B