The pOH of a solution is 8.3. Which of the following is true about the solution?

a
It is acidic and has a pH of 5.7.

b
It is acidic and has a pH of 3.8.

c
It is basic and has a pH of 5.7.

d
It is basic and has a pH of 3.8.

The volume of the balloon after the change is approximately 2.19 L. The gas laws  behaviour of gases by providing relationships between the temperature, moles, volume and pressure

We can use the combined gas law to solve this problem:  (P1V1)/(n1T1) = (P2V2)/(n2T2)   where P is the pressure, V is the volume, n is the number of moles, and T is the temperature.  

We can assume that the temperature and pressure remain constant, so T1 = T2 and P1 = P2.

Therefore, we can simplify the equation to:  (V1/n1) = (V2/n2)  

Substituting the given values, we get:  (V1/0.6400) = (V2/0.3500)  

Solving for V2, we get:  V2 = (0.3500/0.6400) * V1  V2 = 0.5469 * V1  

The volume of the balloon after the change is 0.5469 times the original volume, or:  V2 = 0.5469 * 4.00 L = 2.1876 L  

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molar mass of empirical formula = (20 x 12.01 g/mol) + (24 x 1.01 g/mol) + (1 x 14.01 g/mol) = 324.44 g/mol ratio = 324.44 g/mol / 324 g/mol = 1.001. The molecular formula of quinine is C20H24N.

The active component of cinchona extracts, which have been used for this purpose since before 1633, is utilized as an antimalarial medication. Quinine has been utilized in conventional cold remedies for its use as a mild antipyretic and analgesic.

Plasmodium falciparum malaria is treated with quinine. Malaria is brought on by the parasite Plasmodium falciparum, which enters the body through the red blood cells. Quinine functions by either eliminating the parasite or halting its growth.

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Answer: The answer for this question is 45.9.

Explanation: Since there are numbers with different significant figures, the number or numbers that have less significant figures will be the one you round to. In this case, the numbers in the problem are 32.6,735, and 1035. Since we need to find the numbers that have the least significant figures, the numbers 32.6 and 735 have three significant figures while 1035 has 4 significant figures. Since we need to choose the numbers that have the least significant figures, these numbers are 32.6 and 735 since they have three significant figures. Round your answer to three significant figures and you get 45.9 as your answer.

The mass in of [tex]FeO[/tex] that were produced in the mixture is 8.49g when a sample of 20.00 g iron metal reacts with oxygen gas to form 27.55 g iron oxide mixture.

Given the mass of iron metal = 20.00g

The mass of iron oxide = 27.55g

Iron reacts with oxygen gas to form a mixture of [tex]FeO[/tex] and [tex]Fe2O3[/tex].

The reaction is as follows: [tex]3Fe + 2O2 --- > FeO + Fe2O3[/tex]

The mass of [tex]FeO[/tex] produced in the mixture can be determined using the following equation:

Mass of [tex]FeO[/tex] = Mass of Iron Oxide Mixture – Mass of [tex]Fe2O3[/tex]

Mass of [tex]FeO[/tex] = 27.55 g - Mass of [tex]Fe2O3[/tex]

as we can see from reaction that 3 moles of iron react to give 1 mole of [tex]Fe2O3[/tex] and 1 mole of [tex]FeO[/tex].

The molar mass of Fe (iron) is = 55.85g/mole and molar mass of [tex]Fe2O3[/tex] = 159.69g/mole

3 * 55.85g of Fe produces 1 * 159.69g of [tex]Fe2O3[/tex]

Then 20.00g of iron produces = 20 * 159.69/3 * 55.85 = 19.06g of [tex]Fe2O3[/tex]

Hence mass of [tex]FeO[/tex] produced = 27.55 - 19.06 = 8.49g

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Given that we don't have the equation mentioned in the question, I will assume that the equation is referred to is the following:

2N2(g) + 5O2(g) → 2N2O5(g)

a) If you start with 2 moles of O2, according to the equation, 5 moles of O2 are needed to react with 2 moles of N2. Therefore, the number of moles of N2O5 created would be (2/5) x 2 = 0.8 moles.

b) If you start with 1 mole of O2, according to the equation, 5/2 moles of O2 are needed to react with 2 moles of N2. Therefore, the number of moles of N2O5 created would be (1/(5/2)) x 2 = 0.4 moles.

c) If you start with 1 mole of N2, according to the equation, 5/2 moles of O2 are needed to react with 2 moles of N2. Therefore, the number of moles of N2O5 created would be (5/2) x 1 = 2.5 moles.

d) If you get 6 moles of N2O5 during your reaction, according to the equation, 2 moles of N2O5 are produced for every 5 moles of O2 used. Therefore, the number of moles of O2 needed would be (5/2) x (6/2) = 7.5 moles.

e) If you get 6 moles of N2O5 during your reaction, according to the equation, 2 moles of N2O5 are produced for every 2 moles of N2 used. Therefore, the number of moles of N2 needed would be (2/2) x (6/2) = 3 moles.

Names: The chemical names for the compounds involved in the equation are:

O2: Oxygen gas

N2: Nitrogen gas

N2O5: Dinitrogen pentoxide gas

A chemical reaction between nitrogen and oxygen is represented by the balanced chemical equation 2N2(g) + 5O2(g) 2N2O5(g). The quantities of the reactants and products used in this reaction were questioned.

We may use stoichiometry to determine the mole ratio of O2 to N2O5, which will allow us to calculate how many moles of N2O5 is generated when 2 moles of O2 are consumed.

The equation shows that in order to produce 2 moles of N2O5, 5 moles of O2 must react with 2 moles of N2. The result is the production of (2/5) x 2 = 0.8 moles of N2O5.

Similar calculations can be used to determine how many moles of N2O5 are created when 1 mole of O2 or 1 mole of N2 is used.

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A double replacement is Na2SO4 (aq) + BaCl2 BaSO4 (s) + 2 NaCl (aq) + 2 NaCl (aq). The reaction Na2SO4 + BaCl2 is endothermic. When barium chloride (BaCl2) and sodium sulphate (Na2SO4) combine, sodium chloride and barium sulphate are formed.

When Sodium sulphate(Na 2 SO 4) interacts with Barium chloride solution (), a white precipitate of Barium sulphate() and Sodium chloride is generated.

When barium chloride is introduced to dilute sulphuric acid, a white precipitate of barium sulphate forms as a result of barium displacement from its chloride, as seen below: BaCl2 + H2SO4 BaSO4 + 2HCl.

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a. The rate law rate=k[NO2]^2 indicates that the reaction is second order with respect to NO2.

b. The rate law rate=k indicates that the reaction is zero order with respect to the reactant(s).

c. The rate law rate=k[H2][Br2]^1/2 indicates that the reaction is first order with respect to H2 and half-order with respect to Br2. Therefore, the overall order of the reaction is 1 + 1/2 = 3/2 order.

d. The rate law rate=k[NO]^2[O2] indicates that the reaction is third order overall. The reaction is second order with respect to NO and first order with respect to O2.

The term "rate law" is commonly used to refer to the integrated rate law. K in a rate law is the rate constant, a value specific to each reaction that determines the rate of reaction. Orders in a rate law describe the dependency of the reaction rate on the concentration of each reactant, with each reactant having its own order. The overall reaction order is the sum of the individual orders, which can be determined through experiments.

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Answer: The solubility of NaNO3 (sodium nitrate) in water increases as temperature increases. Therefore, we need to know the solubility of NaNO3 at different temperatures to determine the temperature at which 140 g of NaNO3 can be fully dissolved in water.

According to the solubility curve for NaNO3, the maximum solubility of NaNO3 in water is approximately 88 g per 100 g of water at 80°C. This means that at 80°C, we can dissolve 88 g of NaNO3 in 100 g of water to make a saturated solution.

To fully dissolve 140 g of NaNO3, we need to dissolve it in a sufficient amount of water that can dissolve at least 140 g of NaNO3. Using a proportion, we can calculate the amount of water required to dissolve 140 g of NaNO3 at 80°C:

88 g NaNO3 / 100 g water = 140 g NaNO3 / x g water

Solving for x, we get:

x = 159.1 g water

This means that at 80°C, we need to dissolve 140 g of NaNO3 in at least 159.1 g of water to make a saturated solution, in which all of the NaNO3 will be dissolved.

Therefore, the temperature at which 140 g of NaNO3 can be fully dissolved in water is approximately 80°C.

Explanation:

The heat of reaction at constant pressure when methane is reacted with ammonia to form hydrogen cyanide is 103.6kJ.

Given that Hydrogen cyanide which is a highly poisonous substance can be prepared by the reaction of methane and ammonia as follows:

[tex]CH_4 (g) + NH_3 (g) -- > HCN(g) + 3 H_2 (g)[/tex]

This is obtained through the following steps:

[tex]N_2 (g) + 3 H_2 (g) -- > 2 NH_3 (g)[/tex]; ΔHrxn = -91.8 kJ;

[tex]C(s) + 2 H_2 (g) -- > CH_4 (g)[/tex]; ΔHrxn = -74.9 kJ;

[tex]2 C(s) + H_2 (g) + N_2 (g) -- > 2 HCN(g)[/tex]; ΔHrxn = +270.3 kJ

The heat of reaction at constant pressure is the sum of the individual reactions such that:

[tex]\delta Hrxn = -91.8 kJ + (-74.9 kJ) + (270.3 kJ) = 103.6 kJ[/tex]

The heat of reaction is the total amount of energy released or absorbed when a chemical reaction occurs. It is calculated by adding the enthalpies of the reactants and subtracting the enthalpies of the products. Hence it concludes that when methane and ammonia react to form hydrogen cyanide, 103.6 kJ of energy is released.

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Whereas freezing, condensation, or deposition are exothermic processes, fusion, vaporisation, and  were endothermic processes. Changes in a system's energy occur together every phase transition.

Potential energy is the type of energy that is transforming during a phase change. Either elevated body temperature (PE increases) and released (PE drops) throughout a phase change will permit the molecules either move apart or join together.

Melting is the process of changing from of the solid to liquid state. freezing: the transformation of a liquid into a solid. The process of evaporation is the change from a liquid to a gas. Condensing: The change from a gaseous to a liquid state.

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The partial pressure of CH₄ in the mixture is 0.464 atm.

To find the partial pressure of CH₄ in the mixture, we need to use the following formula:

Partial pressure of CH₄ = Total pressure × mole fraction of CH₄

To calculate the mole fraction of CH₄, we need to first calculate the moles of each gas present in the mixture.

The number of moles can be calculated using the following formula:

moles = mass ÷ molar mass

The molar mass of CH₄ is 16.04 g/mol and the molar mass of Xe is 131.29 g/mol.

Moles of CH₄ = 7.5 g ÷ 16.04 g/mol = 0.467 mol

Moles of Xe = 7.5 g ÷ 131.29 g/mol = 0.057 mol

The total number of moles in the mixture is the sum of the moles of CH₄ and Xe:

Total moles = 0.467 mol + 0.057 mol = 0.524 mol

Now we can calculate the mole fraction of CH₄:

Mole fraction of CH₄ = Moles of CH₄ ÷ Total moles = 0.467 mol ÷ 0.524 mol = 0.891

Finally, we can use the formula for partial pressure to find the partial pressure of CH₄:

Partial pressure of CH₄ = Total pressure × mole fraction of CH₄

Partial pressure of CH₄ = 0.52 atm × 0.891 = 0.464 atm

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The mass of carbon dioxide produced from the complete combustion of 8.40x10^-3 g of methane is 0.023 g.

Mass can be defined as the measure of the amount of matter in a body.

The balanced chemical equation for the complete combustion of methane (CH4) is:

CH4 + 2O2 → CO2 + 2H2O

This equation tells us that one mole of methane reacts with two moles of oxygen gas (O2) to produce one mole of carbon dioxide (CO2) and two moles of water (H2O).

We can use the molar mass of methane and the balanced equation to determine the amount of carbon dioxide produced from the given mass of methane.

First, we need to convert the mass of methane to moles:

moles of CH4 = mass / molar mass = 8.40x10^-3 g / 16.04 g/mol = 5.239x10^-4 moles

Next, we can use the balanced equation to find the number of moles of CO2 produced:

1 mole of CH4 produces 1 mole of CO2

So, 5.239x10^-4 moles of CH4 will produce 5.239x10^-4 moles of CO2.

Finally, we can use the molar mass of carbon dioxide to convert moles to grams:

mass of CO2 = moles of CO2 × molar mass of CO2

mass of CO2 = 5.239x10^-4 moles × 44.01 g/mol = 0.023 g

Therefore, the mass of carbon dioxide produced from the complete combustion of 8.40x10^-3 g of methane is 0.023 g.

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The number of particles in the gas is [tex]4.72 * 10^{24}[/tex], the volume of the gas at 4 atm is 40 L, and the volume of the gas at 300 K is 30 L.

Avogadro's Law: Equal volumes of gases at the same temperature and pressure contain equal numbers of particles (molecules or atoms).

Boyle's Law: For a fixed amount of gas at a constant temperature, the pressure and volume are inversely proportional to each other.

Charles's Law: For a fixed amount of gas at a constant pressure, the volume and temperature are directly proportional to each other.

PV = nRT is the formula for the ideal gas law,

where P is pressure, V is volume, n is the number of molecules of gas, R is gas constant, and T is temperature in Kelvin.

Mass of gas (m) = 4 g

Temperature (T) = 200 K

Pressure (P) = 8 atm

Volume (V) = 20 L

First, we can use the ideal gas law to calculate the number of moles of gas:

n = PV/RT

n = (8 atm * 20 L) / (0.0821 L.atm/mol.K * 200 K)

n = 7.85 moles

Next, we can use Avogadro's Law to find the number of particles (molecules or atoms):

1 mole of gas = [tex]6.02 * 10^23[/tex] particles

7.85 moles of gas =[tex]7.85 * 6.02 * 10^23[/tex]particles

= [tex]4.72 * 10^24[/tex] particles

We can also use Boyle's Law and Charles's Law to find the volume of the gas at different conditions:

Boyle's Law:

[tex]P_1V_1 = P_2V_2[/tex]

If we keep the temperature constant at 200 K, we can use this relationship to find the volume of the gas at a different pressure. Let's say we want to know the volume of the gas at 4 atm:

[tex]P_1[/tex] = 8 atm

[tex]V_1[/tex] = 20 L

[tex]P_2[/tex] = 4 atm

[tex]V_2[/tex]= ?

[tex]P_1V_1 = P_2V_2[/tex]

8 atm x 20 L = 4 atm x [tex]V_2[/tex]

[tex]V_2[/tex] = (8 atm x 20 L) / 4 atm

[tex]V_2[/tex] = 40 L

Charles's Law:

[tex]V1/T1 = V2/T2[/tex]

If we keep the pressure constant at 8 atm, we can use this relationship to find the volume of the gas at a different temperature.

Let's say we want to know the volume of the gas at 300 K:

[tex]V_1[/tex] = 20 L

[tex]T_1[/tex]= 200 K

[tex]V_2[/tex] = ?

[tex]T_2[/tex] = 300 K

[tex]V_1/T_1 = V_2/T_2[/tex]

20 L / 200 K = [tex]V_2[/tex] / 300 K

[tex]V_2[/tex] = (20 L / 200 K) x 300 K

[tex]V_2[/tex] = 30 L

Therefore, the number of particles in the gas is [tex]4.72 * 10^{24}[/tex], the volume of the gas at 4 atm is 40 L, and the volume of the gas at 300 K is 30 L.

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The total enzyme concentration [Et] is 6 nM.

Enzyme concentration is the amount of an enzyme present in a given solution. The concentration of enzymes can have an effect on the rate of reaction. If the concentration of enzymes is higher, the rate of reaction will be faster, and if the concentration of enzymes is lower, the rate of reaction will be slower.Enzyme concentration is important because it can affect the outcome of a reaction, and therefore, it must be carefully monitored.

The total enzyme concentration [Et] can be calculated using the Michaelis-Menten equation, which states that[tex]Vo= [Et] * kcat * (\frac{[S]}{Km} + [S])[/tex]

Plugging in the given values, we get:

[tex]480 nM/min = [Et] * 20 min^{-1}* (\frac{6mM}{4uM} + 6mM)[/tex]

Solving for [Et], we get:

[Et] = 6 nM

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Answer: NaOH (s) → NaOH (aq) is an exothermic process because energy is released when solid NaOH dissolves in water to form an aqueous solution.

NH4Cl (s) → NH4Cl (aq) is also an exothermic process for the same reason as NaOH above.

CaCl2 (s) → CaCl2 (aq) is also an exothermic process because energy is released when solid CaCl2 dissolves in water to form an aqueous solution.

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) is an exothermic process because energy is released as the acid-base reaction occurs and the products are formed.

NaCl (s) → NaCl (aq) is not an endothermic or exothermic process since there is no change in the state of the substance (solid to aqueous solution) and therefore no energy is either absorbed or released.

Explanation: Endothermic and exothermic refer to processes that either require energy to be absorbed or release energy to the surroundings.

In the given reactions, NaOH, NH4Cl, and CaCl2 dissolve in water to form aqueous solutions, and in doing so, they release energy to the surroundings, making these processes exothermic.

The reaction between HCl and NaOH also releases energy as the products are formed, and so it is also exothermic.

However, in the case of NaCl, there is no change in the state of the substance, and so no energy is either absorbed or released, making it neither endothermic nor exothermic.

Hypothesis: Increasing the cooling rate during the solidification of aluminum will result in a finer grain structure in the final solid.

When a liquid metal such as aluminum is cooled and solidified, the atoms in the liquid begin to arrange themselves into a crystalline structure. The rate at which this happens can have a significant effect on the final microstructure of the solid. If the cooling rate is slow, larger grains will form as the atoms have more time to move and arrange themselves into larger clusters. If the cooling rate is faster, there is less time for the atoms to move and larger clusters cannot form, resulting in a finer grain structure. Therefore, the hypothesis proposes that increasing the cooling rate during the solidification of aluminum will lead to a finer grain structure in the final solid. This hypothesis could be tested by varying the cooling rate during the solidification process and then examining the microstructure of the resulting solid.

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Hypothesis: Increasing the cooling rate during the solidification of aluminium will result in a finer grain structure in the final solid.

The following methods can be used to demonstrate whether a solid is:

Ionic (NaCl):

Solubility test: Since NaCl is extremely soluble in water, its ionic nature can be verified by this fact.

Conductivity test: NaCl transmits electricity when it is molten or dissolved because it contains charged ions.

Metallic (Ca):

Conductivity test: Due to the existence of free electrons within their crystal structure, metals like calcium carry electricity.

Ductility and malleability test: Due to their malleability and ductility, metals are readily deformed when under pressure.

Covalent Network (Quartz, SiO2):

Hardness test: Because of the intense covalent bonds between atoms, covalent network solids like quartz are exceedingly hard.

Melting point test: Due to the powerful intermolecular forces between atoms, covalent network solids frequently have high melting and boiling points.

Polar Molecular (sugar, C6H12O6):

Solubility test: Sugar and other polar compounds can dissolve in polar solvents like water but cannot dissolve in nonpolar solvents.

Melting and boiling point test: Due to weaker intermolecular interactions, polar molecular solids have lower melting and boiling points than ionic or covalent network solids.

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The Molecular Equation is : [tex]HI (aq) + KOH (aq) \rightarrow KI (aq) + H_2O (l)[/tex]

The Net Ionic Equation is : [tex]H^+(aq) + OH^-(aq) \rightarrow H_2O(l)[/tex]

This reaction is an acid-base neutralization reaction. Hydroiodic acid (HI) is an acid and potassium hydroxide (KOH) is a base. When the two react, they form water ([tex]H_2O[/tex]) and potassium iodide (KI).[tex]H_2O[/tex] is a chemical formula for water, which consists of two hydrogen atoms and one oxygen atom. Water is a fundamental component of life and is essential for all forms of life on Earth. The reaction is a double displacement reaction, where the positively charged cations, potassium (K⁺) and hydrogen (H⁺), switch places with each other in the reactants to form the products.

It is possible to describe the as:

[tex]HI (aq) + KOH (aq) \rightarrow KI (aq) + H_2O (l)[/tex]

One way to express the net-ionic equation is as follows:

[tex]H^+(aq) + OH^-(aq) \rightarrow H_2O(l)[/tex]

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To solve both problems, we need to use the equations for calculating boiling point elevation and freezing point depression:

Boiling point elevation: ΔTb = Kb x molality

Freezing point depression: ΔTf = Kf x molality

where:

For both problems, we assume that the solute is completely dissolved in the solvent and that the resulting solution is ideal.

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

molality = moles of solute / mass of solvent in kg

Next, we need to find the boiling point elevation and freezing point depression:

ΔTb = Kb x molality

ΔTf = Kf x molality

For water, Kb = 0.512 °C/m and Kf = 1.86 °C/m.

ΔTb = 0.512 °C/m x 1.09 mol/kg = 0.558 °C

ΔTf = 1.86 °C/m x 1.09 mol/kg = 2.03 °C

Therefore, the boiling point of the solution is 100 + 0.558 = 100.558 °C, and the freezing point is 0 - 2.03 = -2.03 °C.

2. First, we need to calculate the molality of the benzoic acid solution:

molality = moles of solute / mass of solvent in kg

Next, we need to find the boiling point elevation and freezing point depression:

ΔTb = Kb x molality

ΔTf = Kf x molality

For water, Kb = 0.512 °C/m and Kf = 1.86 °C/m.

ΔTb = 0.512 °C/m x 0.819 mol/kg = 0.419 °C

ΔTf = 1.86 °C/m x 0.819 mol/kg = 1.52 °C

Therefore, the boiling point of the solution is 100 + 0.419 = 100.419 °C, and the freezing point is 0 - 1.52 = -1.52 °C.

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Answer:

where is the other temperature? so i can help you

Answer: 2.78

Explanation:

 To calculate the pH of 0.16M CH3COOH, we need to use the dissociation constant (Ka) of the acid, which is given as 1.74 × 10^-5 mol dm^-3.

The dissociation of CH3COOH is as follows:

CH3COOH + H2O ↔ CH3COO- + H3O+

The equilibrium constant expression for the dissociation of CH3COOH is:

Ka = [CH3COO-][H3O+] / [CH3COOH]

We can assume that [H3O+] is equal to [CH3COO-] since the acid is weak and will not dissociate completely. Therefore, the equilibrium constant expression can be simplified as:

Ka = [H3O+]^2 / [CH3COOH]

[H3O+]^2 = Ka x [CH3COOH]

[H3O+]^2 = 1.74 × 10^-5 x 0.16

[H3O+]^2 = 2.784 × 10^-6

[H3O+] = √2.784 × 10^-6

[H3O+] = 0.00167 M

Therefore, the pH of the solution can be calculated as:

pH = -log[H3O+]

pH = -log(0.00167)

pH = 2.78

Therefore, the pH of the 0.16M CH3COOH solution is 2.78.

Answer: 0.0499 moles of sodium hydroxide

Explanation:

We can determine the number of moles by using the Molarity equation:

Molarity = Moles / Volume (volume is always in Liters)

We have 2 out of 3 components of this equation

Molarity = 1.000 M

Volume = 49.93 mL / 1000 = 0.0499 L

So we rearrange the equation like the following:

Moles = Molarity x volume

          = 1.000 Mx 0.0499 L

          = 0.0499 moles of sodium hydroxide

Answer:

The answer is [tex]4.94 x 10^23[/tex]  atoms.

To determine the total number of atoms in 62.5 g of carbon disulfide (CS2), we need to use Avogadro's number and the molecular weight of CS2.

The molecular weight of CS2 is:

Carbon atomic weight = 12.01 g/mol

Sulfur atomic weight = 32.06 g/mol

2 sulfur atoms x 32.06 g/mol + 1 carbon atom x 12.01 g/mol = 76.13 g/mol

Now, we can use the formula:

Number of moles = mass ÷ molecular weight

Number of moles of CS2 = 62.5 g ÷ 76.13 g/mol = 0.821 mol

Finally, we can use Avogadro's number, which is[tex]6.022 x 10^23[/tex] atoms per mole, to calculate the total number of atoms in 62.5 g of CS2:

Total atoms = Number of moles × Avogadro's number

Total atoms = 0.821 mol ×[tex]6.022 x 10^23[/tex] atoms/mol

Total atoms =[tex]4.94 x 10^23[/tex] atoms

Therefore, there are approximately[tex]4.94 x 10^23[/tex]atoms in 62.5 g of carbon disulfide (CS2).

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The pOH of the solution is 10.92.

Equation: The relationship between a solution's pH and pOH is:

pH + pOH = 14

where pH is equal to the negative logarithm of the concentration of hydrogen ions (H+), and pOH is equal to the negative logarithm of the concentration of hydroxide ions (OH-).

We must first determine the [OH-] concentration to determine the solution's pOH. We can apply the following:

Kw = [H+][OH-] = 1.0 x 10^-14 at 25°C

where Kw is the water-specific ion product constant.

If we rewrite this equation, we obtain:

[OH-] = Kw/[H+] = 1.0 x 10^-14 / 8.4 x 10^-3 = 1.19 x 10^-11 mol/L

We can now calculate pOH's value using the definition of the term:

pOH = -log[OH-] = -log(1.19 x 10^-11) = 10.92

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The IUPAC name for the given molecule, which has four carbon atoms, one OH group, and two methyl (CH3) groups attached, is:

4-methyl-2-pentanol

What is IUPAC?

It is an international organization that aims to advance the chemical sciences and to contribute to the application of chemistry in the service of humankind.

Identify the longest carbon chain that contains the functional group (OH). In this case, the longest chain has five carbon atoms.

Number the carbon atoms in the chain, starting from the end that is closest to the functional group (OH). In this case, we can number the chain from left to right, so that the OH group is on carbon atom 2.

Identify and name the substituents (groups) attached to the main chain. In this case, there are two methyl (CH3) groups attached to carbon atoms 4 and 5, respectively.

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The flammability of natural gas ranges from 5 to 15 percent. This indicates that no combustion would take place in any mixture with a natural gas to air ratio of less than 5% or greater than 15%.

The lowest concentration of a gas at which combustion can occur is known as the LEL. A reading in%LEL measures the percentage of that LEL value. The LEL, for instance, is 5% by volume for methane. Half of that amount, or 2.5% by volume, is 50% LEL.

Divide the unknown concentration by the LEL specified in the NFPA Handbook to determine the LEL of any gas in air.

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Explanation:

In the case of trans-decalin, there is no steric interaction between the rings because they are at their maximum distance. As a result, the trans conformation is more stable. Another reason is that trans-decalin does not undergo ring flipping. So, it remains rigid and is more stable.

Stearic interactions between the two rings result in a less stable conformation in case of cis-decalin.It undergoes ring flipping so, unstable conformation. (The First image)

Hence, trans-decalin is more stable than cis-decalin.(The second image)

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The mass of water, H₂O produced by the reaction of 1.06 grams of oxygen gas, O₂ is 0.72 grams

The mass of water, H₂O produced by the reaction of 1.06 grams of oxygen gas, O₂ can be obtained as shown below:

The balanced equation for the reaction is given below

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

160 g of oxygen gas, O₂ reacted to produce 108 g of water, H₂O

Therefore,

1.06 g of oxygen gas, O₂ will react to produce = (1.06 × 108) / 160 = 0.72 g of water, H₂O

Thus, the mass of water, H₂O produced from the reaction is 0.72 g

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