4. Diffrection Jtudies involving X-rays, electrons or Neutrons pire informetion about crystillographic properties of Solids. Compare these three techniques with reference to particle energies and types of information that ean be obtained. Which technique is most appropriate for studying surface Crystallography? Which technigue is used to determine mapnetic structure?

Answer:

The compound with the correct formula is;

A. MNO₃

Explanation:

The number of oxidation states in the metal, M = One oxidation state

The formula of the compound formed by the metal, M = MHCO₃

We note that the ion HCO₃⁻, known as hydrogen carbonate has an oxidation number of -1

Similarly nitrate, NO₃⁻ has an oxidation number of -1, therefore, the metal M can form similar compound formed with HCO₃⁻ with nitrate, and we have;

The possible compounds formed by the metal 'M' includes MHCO₃ and MNO₃.

Answer:

because battery have it's own voltage in it's composition

A) The pH of a 0.75 M acetic acid solution, CH3COOH, Ka = 1.8 x 10⁻⁵: The pH is 2.92.

B) The calculated pH of a 0.75 M diethylamine solution, (CH3CH2)2NH, Kb = 7.1 x 10⁻⁴: The pH is 13.71

C) The hydronium ion concentrations can be represented as [H3O+]. the ratio of the hydronium ion concentrations in each solution is 1.59 x 10⁻¹¹.

A) The pH of a 0.75 M acetic acid solution, CH3COOH, Ka = 1.8 x 10⁻⁵: Firstly, we have the formula for Ka:

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

We need to calculate [H3O+] and [CH3COO-].

Let us represent [H3O+] as x and [CH3COO-] as y.

Let [CH3COOH] be

0.75-x: [H3O+][CH3COO-] / [CH3COOH] = 1.8 x 10⁻⁵x

y / (0.75-x) = 1.8 x 10⁻⁵

x = [H3O+]

pH = - log[H3O+]

pH = - log (1.21 x 10⁻³)

The pH is 2.92.

B) The calculated pH of a 0.75 M diethylamine solution, (CH3CH2)2NH, Kb = 7.1 x 10⁻⁴:Given that Kb = 7.1 x 10⁻⁴.We know that

pKb + pKa = pKw (at 25°C)

Now, we have pKa, which is equal to -log Ka. Let us calculate pKa:

pKa = - log(1.8 x 10⁻⁵)

pKa = 4.74

pKw = 14

pKb = pKw - pKa = 14 - 4.74 = 9.26

Let us now use the formula for Kb:

Kb = [BH⁺][OH⁻] / [B]

Kb = [CH3CH2NH3⁺][OH⁻] / [CH3CH2NH2]

Let us represent [CH3CH2NH3⁺] as x and [OH⁻] as y.

Let [CH3CH2NH2] be 0.75-x.

Therefore, we have:

[CH3CH2NH3⁺][OH⁻] / [CH3CH2NH2] = 7.1 x 10⁻⁴x

y / (0.75-x) = 7.1 x 10⁻⁴

y = [OH⁻]

pOH = -log [OH⁻]

pOH = -log (3.53 x 10⁻³)

The pOH is 2.45.Using the formula

pH + pOH = pKw:

2.92 + 2.45 = 14 - pH + pOH2.

92 + 2.45 = 14 - pH + 11.559.37 - 11.55 = pH

pH = - log(1.93 x 10⁻¹⁴)

The pH is 13.71

C) The hydronium ion concentrations can be represented as [H3O+]. In a 0.75 M acetic acid solution:

[H3O+] = 1.21 x 10⁻³

In a 0.75 M diethylamine solution:

[H3O+] = 1.93 x 10⁻¹⁴

The ratio of the hydronium ion concentrations in each solution is:

1.93 x 10⁻¹⁴ / 1.21 x 10⁻³ = 1.59 x 10⁻¹¹.

Therefore, the ratio of the hydronium ion concentrations in each solution is 1.59 x 10⁻¹¹.

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The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.

The partial pressure has been defined as the pressure exerted by each gas in the mixture.

According to the Dalton's law of partial pressure, the total pressure of gas has been the sum of the partial pressure of the gases in the mixture.

The given partial pressure of gases in the mixture has been:

The total pressure of the gaseous mixture has been:

\(P=P_N_2\;+\;P_O_2\;+\;P_A_R\;+\;P_H_e\;+\;P_H\\P=3\;+\;1.80\;+\;0.29\;+\;0.18\;+\;0.10\;\text {atm}\\P=5.37\;\rm atm\)

The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.

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In recent years, the output of ammonia has remained mostly steady. Ammonia output was projected to reach 150 million metric tonnes worldwide in 2022. With over 64.6 million metric tonnes produced, East Asia has the largest ammonia output.

Sodium metal is produced via the Castner method, which involves electrolyzing molten sodium hydroxide at a temperature of around 330 °C. At that temperature, the molten sodium would begin to dissolve in the melt; below that temperature, the liquid would harden.

One mole (or one gramme) of sodium chloride (NaCl) has a molecular weight of 58.44, making 58.44g the molecular weight of one gramme. You may create 1M NaCl by dissolving 58.44g of sodium chloride in a final amount of 1 litre.

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The original volume of HBr that was titrated can be calculated as the ratio of the moles of HBr to its concentration.

To determine the original volume of HBr that was titrated, we can use the concept of stoichiometry and the equation balanced for the neutralization reaction between HBr and KOH.

The balanced equation is:

HBr + KOH → KBr + H₂O

From the balanced equation, we can see that the stoichiometric ratio between HBr and KOH is 1:1. This means that for every mole of HBr, we need an equal number of moles of KOH to complete neutralization.

First, let's determine the moles of KOH used in the titration:

Moles of KOH = 0.500 M × 0.075 L = 0.0375 mol

Since the stoichiometric ratio is 1:1, this also represents the number of moles of HBr that were neutralized.

Now, we can calculate the original volume of HBr using the concentration of the unknown solution:

Moles of HBr = 0.0375 mol

Concentration of HBr = unknown (let's assume it is C mol/L)

Volume of HBr = Moles of HBr / Concentration of HBr = 0.0375 mol / C mol/L

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the reaction will move towards the right to achieve equilibrium

Given information:Initially N2O5 concentration is at 1.50 atm and the concentration of N2O3 is 5.00 atm and the concentration of O2 is 4.85 atm.N2O5(g) ⇌ 2NO2(g) + 1/2O2(g)

The direction of reaction at equilibrium:When a reaction moves in the forward direction, products are formed from reactants. The concentration of reactants decreases as the concentration of products increases. In contrast, if a reaction moves in the reverse direction, products break down into reactants. The concentration of products decreases as the concentration of reactants increases.In this case, as the reaction is exothermic, the concentration of N2O5 will decrease, and the concentration of NO2 and O2 will increase. Therefore, the reaction will move towards the right side to achieve equilibrium.

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. In this reaction, N2O5(g) dissociates into 2NO2(g) and 1/2O2(g). The reaction is exothermic since it releases energy in the form of heat. Initially, the concentration of N2O5 is 1.50 atm, while the concentrations of NO2 and O2 are 0 atm.

According to Le Chatelier's principle, if there is a change in the reaction conditions, the system will try to oppose that change and re-establish equilibrium.The concentration of NO2 and O2 will increase as the concentration of N2O5 decreases, resulting in a shift in the forward direction. As the reaction proceeds, the concentration of NO2 and O2 will continue to increase, while the concentration of N2O5 will continue to decrease until equilibrium is established. As a result, the reaction will move to the right to reach equilibrium.

the reaction will move towards the right to achieve equilibrium. The concentration of N2O5 will decrease, and the concentration of NO2 and O2 will increase. Le Chatelier's principle predicts the direction of the reaction at equilibrium by analyzing the changes in the reaction conditions. When the concentration of N2O5 decreases, the reaction will shift to the right to re-establish equilibrium.

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Hydrazine is a chemical compound with the formula \(N_2H_4\). 120 molecules of\(NH_3\) will be formed if 60 molecules of \(N_2H_4\) decompose into \(N_2\) and \(NH_3\).

It is used as a rocket fuel and as a polymerization catalyst in the production of plastics and when hydrazine decomposes, it produces nitrogen gas (\(N_2\)) and ammonia (\(NH_3\)). If 60 molecules of hydrazine decompose into \(N_2\) and \(NH_3\), the number of molecules of \(NH_3\) that will be formed can be determined using the balanced equation for the reaction: \(N_2H_4 -- > N_2 + 2NH_3\). For every one molecule of hydrazine that decomposes, two molecules of ammonia are formed. Therefore, the number of ammonia molecules produced is twice the number of hydrazine molecules that decompose. Since 60 molecules of hydrazine are decomposing, the number of ammonia molecules formed is: 2 x 60 = 120 molecules of \(NH_3\)

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The grams are in the 6.95 × 10²⁴ molecules of the SF₆ is 0.0790 grams.

The number of the molecules of SF₆ =  6.95 × 10²⁴ molecules

The molar mass of the SF₆  = 146 g/mol

The one mole of the substance =  6.022 × 10²³ molecules

The number of the moles in the 6.95 × 10²⁴ molecules = 6.95 × 10²⁴ /  6.022 × 10²³

The number of the moles = 11.54 moles

The grams in the 11.54 moles = 11.54 / 146

                                                 = 0.0790 grams

Thus, the 0.0790 grams present in the 6.95 × 10²⁴ molecules of the SF₆ .

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Answer:

The answer is in my attachment

Explanation:

hope this helped!

Answer:

6.93

Explanation:

(0.07)6 + (0.93)7 = 0.42 + 6.51 = 6.93

The major difference between altair and the sun is their difference in mass. They differ in classifications due to variations in age with Altairs being younger compared to The Suns' maturity as well as holding an A-type categorization accredited by elevated mass quantities surpassing those of The Suns'.

While there exist detectable disparities between Altair and the Sun's physical properties respective to each other; there exists a remarkable similarity between them concerning their fundamental building blocks- chemical elements like hydrogen, helium, oxygen, carbon and nitrogen that were developed through an occurrence called nuclear fission inside star cores.

Nuclear fusion is the process by which two or more atomic nuclei come together to form extensive nucleus resulting in massive amounts of energy generation required to power the stars.

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Methanol is a hazardous substance that can cause a range of   problems if ingested or inhaled.

Some of the common hazards associated with methanol include blindness, respiratory failure, and even death. However, one hazard that is not typically associated with methanol is skin irritation or corrosion. While methanol can be harmful if it comes into contact with the skin, it is not known to cause significant irritation or corrosion in the way that some other chemicals do .Methanol is a toxic substance that is commonly used in industrial applications such as fuel, solvents, and antifreeze. The substance is highly flammable and can cause a range of health problems if ingested or inhaled. Methanol poisoning can lead to symptoms such as dizziness, nausea, vomiting, and even death. However, one hazard that is not commonly associated with methanol is skin irritation or corrosion. Unlike some other hazardous substances, methanol is not known to cause significant skin irritation or corrosion. However, it is important to note that methanol can still be harmful if it comes into contact with the skin. If methanol is spilled on the skin, it should be immediately washed off with soap and water. If methanol is ingested, medical attention should be sought immediately.

methanol is a hazardous substance that can cause a range of health problems if ingested or inhaled. While skin irritation or corrosion is not typically associated with methanol, it is still important to take precautions when handling the substance to avoid any potential harm. If you suspect that you have been exposed to methanol, seek medical attention immediately.

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Considering the Dalton's law, the pressure of the pure oxygen is 736.2 torr.

The pressure exerted by a particular gas in a mixture is known as partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

\(P_{T} =P_{1} +P_{2} +...+P_{n}\)

where n is the amount of gases present in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case, you  know that there is a mixture of oxygen gas and gaseous water (steam). Then the total pressure of a gas mixture is

\(P_{T} =P_{O_{2} } +P_{water}\)

You know:

Replacing:

\(760 torr=P_{O_{2} } +23.8 torr\)

Solving:

\(P_{O_{2} } =760 torr - 23.8 torr\)

\(P_{O_{2} } =\)736.2 torr

Finally, the pressure of the pure oxygen is 736.2 torr.

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The correct Greek letter used for the carbon adjacent to carboxyl group is alpha (α).

What is carboxylic acid?

Any of a group of organic compounds known as carboxylic acids in which a carbon (C) atom forms a double bond with an oxygen (O) atom and a single bond with a hydroxyl group (OH). The carbon atom is connected to a hydrogen (H) atom or another univalent combining group by a fourth bond. The carbonyl group (C=O) and hydroxyl group are what give the carboxyl (COOH) group its name. The acidity of the carboxylic acids is one‘s primary chemical characteristic. They are typically weaker than the well-known mineral acids but generally more acidic than other organic compounds with hydroxyl groups (e.g., hydrochloric acid, HCl, sulfuric acid, H2SO4, etc.).

Greek letter used for the carbon adjacent to the carboxyl group is Alpha.

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Answer:

This is probably too late but the answer is 351

Explanation:

0.73x751=548.23

After getting both new pressure and new volume and having only one original pressure which is 1.56 you´ll have to divide 548.23 by 1.56 and you´ll end up with 351 which should be the original volume.

Molarity is the molar concentration of the solute dissolved in a volume of a solution. The molarity of the solution prepared by dissolving barium chloride will be 0.085 M.

Molarity is the ratio of the moles of the solute to that of the volume of the solution in Liters. It can be given as,

\(\rm Molarity = \dfrac{moles}{Volume}\)

Here, moles of the barium chloride can be given by the mass and the molar mass and volume is given as 0.450 L.

Substituting values in the equation:

\(\begin{aligned}\rm Molarity &= \rm \dfrac{mass}{Molar \;mass\times Volume}\\\\&= \dfrac{8}{208.23\times 0.450}\\\\&= 0.085\;\rm M\end{aligned}\)

Therefore, 0.085 M barium chloride is the molar concentration.

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Answer:

Hydration

Explanation:

* 2-propanol is converted to propene using heated alumina beads as the catalyst. The product gas is purified with a dry ice–propanol cold trap.

* the propene is hydrogenated to propane using a palladium catalyst in nearly 100% yield

Answer:

PbI₃ is the precipitate in the reaction

Explanation:

A precipitate is defined as a substance than appears after a chemical reaction because of the formation of an insoluble compound.

You can identify yhe precipitate in a chemical reaction because its state is solid (s). For example, in the reaction:

Pb(NO₃)₃(aq) + 3KI(aq) → PbI₃(s) + 3KNO₃(aq)

The Pb(NO₃)₃ and KI are soluble salts (Because are in (aq) medium) but the reaction produce a solid, PbI₃ and a soluble salt, KNO₃. The solid,

Answer:

See explanation

Explanation:

Ga is in group 13 hence it must loose three electrons to form Ga^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

O is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form O^2- since oxygen has six electrons on its outermost shell.

Br in group 17 has seven electrons in its outermost shell hence it must form Br^- (gain one electron) in order to attain the noble gas configuration.

P in group 15 must accept three electrons and form P^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

S is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form S^2- since sulphur has six electrons on its outermost shell.

Mg in group 2 has two electrons on its outermost shell and must loose both to attain the noble gas configuration forming Mg^2+.

Al is in group 13 hence it must loose three electrons to form Al^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

Se is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form Se^2- since selenium has six electrons on its outermost shell.

Lithium is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Li^+.

Rb is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Rb^+.

As in group 15 must accept three electrons and form As^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

I in group 17 has seven electrons in its outermost shell hence it must form I^- (gain one electron) in order to attain the noble gas configuration.

Aqueous potassium permanganate that has been diluted with carbon tetrachloride with bromine. Pentane and 1-pentane are distinguished by reagents.

An unsaturation test can be performed using the bromine in carbon tetrachloride solution. Use 100 ml of carbon tetrachloride and bromine to evaluate the unsaturation of a compound when it is insoluble in water. The bromine water turns gray as a result of how the bromine reacts with an alkene's carbon-carbon double bonds. With the help of bromine, an alkene is transformed into an alkane by breaking the carbon-carbon bond.

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The overall reaction for the given cell line notation of a galvanic cell is:

B. \(2Al(s) + 6H+(aq)\) → \(2Al_3+(aq) + 3H_2(g)\)

The given cell line notation represents a galvanic cell consisting of two half-cells. On the left side, we have the aluminum electrode (Al(s)) in contact with a solution of AP+ ions (AP+(aq)), while on the right side, we have a hydrogen electrode (\(H_2\)(g)) in contact with an acidic solution (H+(aq)) and a platinum electrode (Pt(s)).

The balanced reaction in the galvanic cell is represented by the overall cell line notation. By examining the notation, we can see that aluminum (Al) is oxidized, losing electrons to become \(Al_3\)+ ions, while hydrogen ions (H+) from the acidic solution are reduced, gaining electrons to form hydrogen gas (\(H_2\)). The presence of the platinum electrode (Pt(s)) serves as a catalyst and does not participate in the overall reaction.

In summary, the overall reaction for the given galvanic cell line notation is 2Al(s) + 6H+(aq) → 2Al3+(aq) + 3\(H_2\)(g), as mentioned in option B.

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If you added 0.5 g of sample 1 instead of 2.0 g, the freezing point temperature of the solution would decrease.

When a solute is added to a solvent, it disrupts the formation of the solvent's crystal lattice structure, lowering the freezing point of the solution. The extent to which the freezing point is lowered depends on the concentration of the solute particles in the solution. In this case, by reducing the amount of sample 1 from 2.0 g to 0.5 g, the concentration of solute particles in the solution would decrease.

Since the freezing point depression is directly proportional to the concentration of solute particles, a decrease in the amount of sample 1 would result in a smaller decrease in the freezing point temperature compared to if 2.0 g were added. In other words, the solution would experience a less significant decrease in freezing point temperature with only 0.5 g of sample 1.

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Answer:

That's is true maybe I'm not sure check your book

The substance that is the oxidizing agent in the reaction is E.\(PbO2\)

A chemical species known as an oxidizing agent has a propensity to oxidise other compounds. This indicates that it causes a material to become more oxidised by causing it to lose electrons. Normal oxidising agents live in their most oxidized forms and have a significant propensity to pick up electrons and proceed through reduction. Strongly electron-affine ions, atoms, and molecules are regarded as effective oxidizers. The power of oxidation increases with increasing electron affinity.

Pb is the chemical that undergoes a reduction in the process, changing its oxidation state from 0 to +2 in the formation of \(PbSO4\). The oxidising agent is the chemical that results in the reduction of Pb. \(PbO2\) serves as the oxidising agent in this situation because it contributes oxygen atoms to the oxidation process.

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a) In a Zero-order reaction after 18 months, the drug concentration in the nebulizer is 55 mg/mL. b) In a First-order reaction after 18 months, the drug concentration in the nebulizer is 59.2 mg/mL.

a) For a zero-order reaction, the rate of drug degradation is constant and independent of the drug concentration. Therefore, the rate equation can be written as:

Rate = k

where k is the rate constant.

Using the given data, we can calculate the rate constant as:

Rate = -Δ[C]/Δt = -(75 - 100)/(10 months) = 2.5 mg/ml/month

So, the rate constant (k) is 2.5 mg/mL/month.

The half-life of a zero-order reaction can be calculated using the following equation:

t1/2 = [C0] / (2k)

where [C0] is the initial concentration of the drug.

Plugging in the given values, we get:

t1/2 = 100 mg/mL / (2 × 2.5 mg/mL/month) = 20 months

Therefore, the half-life of the drug in the nebulizer is 20 months.

To calculate the amount of drug left after 18 months, we can use the following equation:

[C] = [C0] - kt

where [C] is the concentration of the drug at time t, [C0] is the initial concentration of the drug, k is the rate constant, and t is the time interval.

Plugging in the given values, we get:

[C] = 100 mg/mL - 2.5 mg/mL/month × 18 months = 55 mg/mL

b) For a first-order reaction, the rate of drug degradation is proportional to the drug concentration. Therefore, the rate equation can be written as:

Rate = k[C]

where k is the rate constant and [C] is the concentration of the drug.

Using the given data, we can calculate the rate constant as:

Rate = -Δln[C]/Δt = ln([C0]/[C])/Δt = ln(100/75)/(10 months) = 0.0301 1/month

So, the rate constant (k) is 0.0301 1/month.

The half-life of a first-order reaction can be calculated using the following equation:

t1/2 = ln(2) / k

Plugging in the given values, we get:

t1/2 = ln(2) / 0.0301 1/month = 23.0 months

Therefore, the half-life of the drug in the nebulizer is 23.0 months.

To calculate the amount of drug left after 18 months, we can use the following equation:

[C] = [C0] × e^(-kt)

where [C] is the concentration of the drug at time t, [C0] is the initial concentration of the drug, k is the rate constant, and t is the time interval.

Plugging in the given values, we get:

[C] = 100 mg/mL × e^(-0.0301 1/month × 18 months) = 59.2 mg/mL

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Answer:

Willow is correct because as things age and mature more, usually the more better they are.

Explanation:

To elaborate, you can perceive the development of the modern day atomic model as one aging and maturing. There's usually a correlation between age/maturity and knowledge, the same is true for the atomic model and other scientific theories.