A mixture of He
, N2
, and Ar
has a pressure of 13.6
atm at 28.0
°C. If the partial pressure of He
is 1831
torr and that of Ar
is 997
mm Hg, what is the partial pressure of N2
?

Answer : The molarity of the resulting solution is, 0.29 M

Explanation :

When NaCl dissolved in water then it dissociates to give sodium ions and chloride ions.

Given,

Moles of NaCl = 0.87 mol

Volume of solution = 3.00 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

[tex]\text{Molarity}=\frac{\text{Moles of }NaCl}{\text{Volume of solution (in L)}}[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Molarity}=\frac{0.87mol}{3.00L}=0.29mole/L=0.29M[/tex]

Therefore, the molarity of the resulting solution is, 0.29 M

Answer:

A the reaction will form

Explanation:

Answer:

B) 1-propanol has more hydrogen bonding than ethyl methyl ether.

Explanation:

1-propanol and ethyl methyl ether look very similar but have very different boiling points. They both have three carbon atoms but ethyl methyl ether has an oxygen atom.

Ethyl methyl ether has both London dispersion forces and dipole-dipole interactions.

1-Propanol has London dispersion forces, dipole-dipole interactions and hydrogen bonding.

1-propanol has more hydrogen bonding than ethyl methyl ether which explains why they both have different boiling points.

Final answer:

Only statement b, which says that chlorine is reduced from +7 to -1 in both reactions, is correct. Statements a, c, and d cannot be confirmed as correct based on the given information.

Explanation:

The student has presented two chemical reaction equations involving the thermal decomposition of NH4ClO4 (ammonium perchlorate) and powdered aluminum. Let's address the statements provided:

a. All nitrogen atoms lose three electrons in both reactions. This statement is incorrect. In the provided reactions, nitrogen goes from an oxidation state of -3 in NH4+ to 0 in N2, which means each nitrogen atom gains three electrons.

b. Chlorine is reduced from +7 to -1 in both reactions. This statement is correct. In NH4ClO4, chlorine starts with an oxidation state of +7 and is reduced to -1 in HCl.

c. Reaction 2 produces more energy per gram of Al. Without details of the mass of aluminum involved in Reaction 2, we cannot determine which reaction produces more energy per gram of aluminum. However, if the reactions involve the same mass of aluminum, then Reaction 1 is more energetic since the absolute value of ΔH is greater.

d. The thrust produced by the formation of gaseous products is greater per mole of ammonium perchlorate in Reaction 2. This statement cannot be evaluated without more information about the moles of gaseous products formed in Reaction 2.

Therefore, based on the information provided, the correct statement is b. Chlorine is reduced from +7 to -1 in both reactions.

Answer:

7.26 moles of NH₃ are formed in this reaction

Explanation:

This is about the reaction for the production of ammonia

1 mol of nitrogen gas reacts to 3 moles of hydrogen in order to produce 2 moles of ammonia.

The equation is: N₂ + 3H₂ → 2NH₃

In the question, we were informed that the excess is the H₂ so the N₂ is limiting reagent. We determine the moles, that has reacted:

101.7 g / 28 g/mol = 3.63 moles

So, If 1 mol of nitrogen gas can produce 2 moles of ammonia

3.63 moles of N₂ must produce ( 2 . 3.63) / 1 = 7.26 moles of NH₃

Answer:

In this reaction, 7.26 moles of NH3 will be formed.

Explanation:

Step 1: Data given

Mass of N2 = 101.7 grams

Molar mass N2 = 28.0 g/mol

H2 is in excess

Molar mass H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2/ molar mass N2

Moles N2 = 101.7 grams / 28.0 g/mol

Moles N2 = 3.63 moles

Step 4: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 3.63 moles N2 we'll produce2*3.63 = 7.26 moles NH3

In this reaction, 7.26 moles of NH3 will be formed.

Answer:

We should weigh 72.1 of sodium sulfate

Explanation:

The solution must be made of sodium sulfate. This salt can be dissociated like this:

Na₂SO₄ →  2Na⁺  +  SO₄⁻²

From this dissociation we can say that, 2 moles of sodium cation are obtained from 1 mol of salt.

Therefore 0.513 moles of sodium cation will be obtained from 0.2565 moles of salt (0.513 . 1) / 2. The thing is that this number are the moles contained in 1 L of solution, and we need to prepare such 1.98 L

Molarity = mol / Volume (L) → Molarity . volume (L) = mol

0.2565 mol/L . 1.98L = 0.508 moles

These are the moles we should weigh. Let's convert them to moles:

0.508 mol . 142.06 g / 1 mol = 72.1 g

Answer:

We need 72.1 grams of Na2SO4

Explanation:

Step 1: Data given

sodium ion concentration = 0.513 mol/L

Volume = 1.98 L

Step 2: The balanced equation

Na2SO4 → 2Na+ + SO4^2-

Step 3: Calculate moles Na+

Moles Na+ = molarity Na+ * volume

Moles Na+ = 0.513 * 1.98 L

Moles Na+ = 1.01574 moles

Step 4: Calculate moles Na2SO4

For 1 mol Na2SO4 we need 2 moles Na+ and 1 mol SO4^2-

For 1.01574 moles Na+ we'll need 1.01574/2 = 0.50787 moles Na2SO4

Step 5: Calculate moles Na2SO4

Mass Na2SO4 = moles Na2SO4 * molar mass Na2SO4

Mass Na2SO4 = 0.50787 moles * 142.04 g/mol

Mass Na2SO4 = 72.1 grams

We need 72.1 grams of Na2SO4

Answer:

Humid Continental

Marine west coast

Mediterranean

subarctic

Explanation:

just did assignment on edge

Marine west coast and Mediterranean are the types of temperate climates, due to the dispersion of precipitation throughout the year, temperate marine climates are typically distinguished by a notable lack of dry season, hence options D and E are correct.

Temperate climates are regions with moderate annual or seasonal rainfall, intermittent drought, mild to warm summers, and cool to cold winters.

Humid subtropical, marine west coast, Mediterranean are the phrases that clearly identified with temperate marine climates.

Geographically speaking, the moderate climates of Earth are found in the middle latitudes, which are halfway between the tropics and the poles.

Therefore, options D and E are correct.

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Answer:

V= 9.45L

Explanation:

P=3.75atm, V=?, n= 1.5, R= 0.082, T= 15+ 273= 288K

Applying

PV= nRT

Substitute and Simplify

3.75*V= 1.5*0.082*288

V= 9.45L

The approximate volume of a 1.50 mol sample of gas at 15.0°C and a pressure of 3.75 atm is calculated using the Ideal Gas Law to be about 9.20 liters.

To calculate the approximate volume of a 1.50 mol sample of gas at 15.0°C and a pressure of 3.75 atm, we can use the Ideal Gas Law, which is PV = nRT. First, we must convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. Then we can solve for V, the volume of the gas.

The conversion from Celsius to Kelvin: T(K) = 15.0 + 273.15 = 288.15 K

Using the Ideal Gas Law constants, R = 0.0821 L·atm/K·mol. Substituting the known values into the Ideal Gas Law equation:

PV = nRT
(3.75 atm) × V = (1.50 mol) × (0.0821 L·atm/K·mol) × (288.15 K)

V = (1.50 mol × 0.0821 L·atm/K·mol × 288.15 K) / 3.75 atm

V = 9.2029 L

Therefore, the approximate volume of the gas sample is about 9.20 liters.

In atoms, electrons exist on specific, discrete energy levels, which correspond to distinct values of the principal quantum number 'n'. Electricity fills these 'shells' from the lowest to the highest energy state. Transitions between energy levels require definite energy changes.

The energy levels of electrons in an atom can be understood using the principal quantum number often denoted by 'n'. Electrons exist only on specific, discrete energy levels and not between them. This principle is known as the quantization of energy. Each energy level is associated with a specific value of 'n', with n = 1, 2, 3, and so on. The greater the 'n' value, the higher the energy level.

Each energy level or shell can be visualized as concentric circles radiating from the nucleus of the atom. Electrons fill these shells starting from the one closest to the nucleus (lowest energy state), filling up to the one furthest from the nucleus (highest energy state). This illustrates the principle of energy states in a metal as in Figure 9.13.

Furthermore, the Bohr atom model suggests that each electron orbit around the nucleus corresponds to a distinct energy level, and each transition between these levels involves a definite energy change. Therefore, the energy levels of electrons in an atom are directly related to the amount of energy possessed by the electrons.

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Answer:

9.12x10¹³ Hz

Explanation:

The vibrational frequency (ω) of a molecule is given by:

[tex] \omega = \sqrt{\frac{k}{\mu}} [/tex]

Where:

k: is the spring constant

μ: is the reduced mass

The reduced mass of a diatomic molecule is:

[tex] \frac{1}{\mu} = \frac{1}{m_{a}} + \frac{1}{m_{b}} [/tex]

Where ma and mb are the atomic masses of the atoms a and b, respectively, of the diatomic molecule.

Hence, the vibrational frequency of the hydrogen molecule is:

[tex]\omega_{H_{2}} = \sqrt{\frac{k}{\mu_{H_{2}}}}[/tex]   (1)

From equation (1) we can find k:

[tex] k = \omega_{H_{2}}^{2}*\mu_{H_{2}} [/tex]    (2)

The vibrational frequency of the deuterium molecule is:

[tex] \omega_{D_{2}} = \sqrt{\frac{k}{\mu_{D_{2}}}} [/tex]    (3)

By entering equation (2) into equation (3) we can calculate the vibrational frequency of the deuterium molecule:

[tex] \omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{\mu_{D_{2}}}} [/tex]

[tex] \omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{2*\mu_{H_{2}}}} [/tex]

[tex] \omega_{D_{2}} = \frac{\omega_{H_{2}}}{\sqrt{2}} = \frac{1.29 \cdot 10^{14} Hz}{\sqrt{2}} = 9.12 \cdot 10^{13} Hz [/tex]

Therefore, the vibrational frequency of the deuterium molecule is 9.12x10¹³ Hz.

I hope it helps you!

The vibrational frequency of D₂ is :   9.12 * 10¹³ Hz

Given that:

Vibrational frequency ( w ) = [tex]\sqrt{\frac{k}{u} }[/tex]

u = reduced mass

The reduced mass of a diatomic molecule is expressed as

[tex]\frac{1}{u} = \frac{1}{m_{a} } + \frac{1}{m_{b} }[/tex]

Where : Ma and Mb are the atomic masses of mass A and mass B

First step : expressing the vibrational frequency of the hydrogen molecule

wH₂ = [tex]\sqrt{\frac{k}{uH_{2} } }[/tex]   ----- ( i )

from the equation

k = ( wH₂ )² * uH₂ ---- ( ii )

Next step : expressing the vibrational frequency of the deuterium molecule.

wD₂ = [tex]\sqrt{\frac{k}{uD_{2} } }[/tex]  ---- ( iii )

Insert equation ( ii ) into equation ( iii )

wD₂ = [tex]\frac{wH_{2} }{\sqrt{2} }[/tex]  = ( 1.29 * 10¹⁴ ) / ( √2 ) = 9.12 * 10¹³ Hz

Hence we can conclude that The vibrational frequency of D₂ is :   9.12 * 10¹³ Hz.

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The dissolution of potassium hydroxide (KOH) in water is an exothermic process, as indicated by the increase in temperature when KOH is dissolved in water.

The dissolution of potassium hydroxide (KOH) in water is an exothermic process. This can be determined based on the observation that the temperature of the water increased from 23.0°C to 34.0°C when 11.1 g of KOH was dissolved in 250.0 g of water. The positive change in temperature indicates that heat was released by the KOH dissolving in water, resulting in an increase in temperature.

The reaction is exothermic as the temperature of the water rises from 23.0°C to 34.0°C, indicating the release of heat.

To determine if the reaction is exothermic or endothermic, we need to look at the temperature change during the dissolution of potassium hydroxide (KOH).

Since the temperature of the water rises from 23.0 °C to 34.0 °C, which is an increase of 11.0 °C, this indicates that the solution absorbs heat.

This increase in temperature demonstrates that the reaction releases heat into the surroundings, thus it is an exothermic reaction. In an exothermic reaction, the temperature of the surroundings rises because energy is released.

Example Calculation:

Mass of water (m): 250 g

Specific heat capacity (C) of water: 4.184 J/g°C

Temperature change (ΔT): 34.0 °C - 23.0 °C = 11.0 °C

Heat (q) absorbed by the solution: q = m × C × ΔT = 250 g × 4.184 J/g°C × 11.0 °C = 11, 506 J or 11.506 kJ

Since the heat is released by the dissolution of KOH, the reaction is said to be exothermic.

Answer:

It would evaporate.

Explanation:

If you left it on room temperature  it would evaporate.

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

The observed rate constant k is related to the individual rate constants of the mechanism by the equation k = k2 (k1/k-1), where k2 is the rate constant of the rate-determining slow step and k1/k-1 is the equilibrium constant of the first fast equilibrium step.

The relationship between the observed rate constant k (in the rate law) and the rate constants for the individual steps (k1, k-1, k2, k3) in the proposed mechanism for the decomposition of nitramide can be determined by examining the rate-determining step (RDS). In a reaction mechanism, the slowest step controls the overall reaction rate. For the given mechanism:

the observed rate law is rate = k [O2NNH2] [H+]. Because the second step is slow, it is the RDS. The equilibrium of the first step means that the concentration of the intermediate O2NNH−(aq) can be expressed in terms of the concentrations of the reactants O2NNH2 and H+. Therefore, the observed rate constant k is a function of the rate constants of the individual steps, particularly k2 and the equilibrium constant (K = k1/k−1) from the first step. Hence, we can conclude that k is equal to k2 multiplied by the equilibrium constant of the first step, where k = k2 (k1/k−1)

Answer : The electronic geometry and the molecular geometry of the molecule will be octahedral.

Explanation :

Formula used  :

[tex]\text{Number of electron pair}=\frac{1}{2}[V+N-C+A][/tex]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

The given molecule is, [tex]PF_6^-[/tex]

[tex]\text{Number of electrons}=\frac{1}{2}\times [5+6+1]=6[/tex]

The number of electron pair are 6 that means the hybridization will be [tex]sp^3d^2[/tex] and the electronic geometry and the molecular geometry of the molecule will be octahedral.

The correct number of regions of electron density around the central atom in PF₆- is 6. The electronic geometry of PF6- is octahedral, and the molecular geometry is also octahedral.

To determine the number of electron regions around the central phosphorus atom in PF6-, we count the bonded atoms and any lone pairs on the central atom.

The Lewis structure of PF6- shows that the phosphorus atom is bonded to six fluorine atoms and has no lone pairs since it is surrounded by six bonding groups and has a formal charge of -1, which balances the overall charge of the ion.

Each bonded fluorine contributes one region of electron density, and there are no lone pairs on the phosphorus atom.

Therefore, there are six regions of electron density around the central phosphorus atom, which corresponds to an octahedral electronic geometry.

 Since there are no lone pairs, the molecular geometry is the same as the electronic geometry, which is octahedral. This means that all six fluorine atoms are arranged around the central phosphorus atom in a way that they are at the vertices of an octahedron.

 In summary, the VSEPR theory predicts that PF6- has six regions of electron density, an octahedral electronic geometry, and an octahedral molecular geometry.

Answer:

Molality = 6.0 m

Explanation:

The molecular weight of tartaric acid = 150.087 g/mol

Given that:

Density  of the solution = 1.023 g/mL

Molarity =  3.23M

Density is given as : [tex]Molarity ( \frac{1}{molality } +\frac{mol.wt}{1000} )[/tex]

∴

[tex]1.023 = 3.23 (\frac{1}{molality } +\frac{150.087}{1000} )[/tex]

[tex]\frac{1}{molality } =( \frac{1.023}{3.23} - \frac{150.087}{1000} )[/tex]

[tex]\frac{1}{molality } =0.3167 - 0.1500[/tex]

[tex]\frac{1}{molality } = 0.1667[/tex]

Molality  = [tex]\frac{1}{0.1667}[/tex]

Molality = 5.999 m

Molality ≅ 6.0 m

The Organisms that formed them likely lived during the Paleozoic era.

Final answer:

When trilobite fossils are found alongside other species in the same rock layer, a scientist can infer that these organisms lived during the same time period and may have had similar environmental conditions. Trilobites are used as index fossils which can determine the age of the rocks they are in due to their distinct presence in the geological record from 500 to 600 million years ago.

Explanation:

When a scientist discovers fossilized remains of other species in the same rock layer as a trilobite, they can draw some important conclusions. Since trilobites were widespread marine animals that lived between 500 and 600 million years ago, any other fossils found within the same stratum are likely from the same geological period. This is due to the principle of superposition, which states that in any undisturbed sequence of rocks deposited in layers, the youngest layer is on top and the oldest on bottom, each layer being younger than the one beneath it and older than the one above it.

Furthermore, trilobites are known as index fossils because they were widespread, rapidly evolving, and limited in geological time. Therefore, they serve as indicators of the age of the rock layer in which they are found. Discovering other species in conjunction with trilobite fossils suggests that these organisms also lived during the same time period and may have shared similar environments. This can be used to infer that these species are from the same age and geological period as the trilobites.

The study of these fossils can provide invaluable insight into the biodiversity and ecological conditions of prehistoric marine environments. Moreover, these findings contribute to our understanding of evolutionary history and may help to identify patterns of mass extinction or other significant pale ontological events.

Answer: This reaction is never spontaneous

Explanation:

According to Gibbs equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibb's free energy change

[tex]\Delta H[/tex] = enthalpy change

T = temperature

[tex]\Delta S[/tex] = entropy change

A reaction becomes spontaneous when [tex]\Delta G[/tex] = Gibb's free energy change is negative.

[tex]\Delta G=+ve-T(-ve)[/tex]

[tex]\Delta G=+ve+ve[/tex]

[tex]\Delta G=+ve[/tex]

Thus this reaction is never spontaneous

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate =[tex]3.45 g\times \frac{32.5}{100}=1.121 mol[/tex]

Moles of bicarbonate ion = [tex]\frac{1.121 g/mol}{61 g/mol}=0.01840 mol[/tex]

[tex]HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)[/tex]

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

[tex]\frac{1}{1}\times 0.01840 mol=0.01840 mol[/tex] of carbon dioxide gas

Moles of carbon dioxide gas  n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

[tex]PV=nRT[/tex] (ideal gas equation)

[tex]V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L[/tex]

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

Final answer:

The volume of CO2 generated from a 3.45-g Alka-Seltzer tablet containing 32.5% bicarbonate is approximately 469.2 mL when reacted with stomach acid at 37°C and 1.00 atm.

Explanation:

To calculate the volume of CO2 generated from the reaction of bicarbonate (HCO^{3−}) with HCl in the stomach, we can use the provided mass percent of bicarbonate in the Alka-Seltzer tablet and the reaction stoichiometry. The balanced equation for the reaction between sodium bicarbonate and hydrochloric acid is:

NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

Assuming the percent by mass of bicarbonate is 32.5%, a 3.45-g tablet would contain 1.12125 g of HCO3−. The molar mass of HCO^{3−} is approximately 61 g/mol, which corresponds to 0.0184 moles of HCO^{3−}. Since the reaction produces one mole of CO2 for each mole of HCO3−, we can use the ideal gas law to calculate the volume of CO2.

At STP (0°C and 1 atm), one mole of any gas occupies 22.4 L. However, we're given conditions of 37°C and 1.00 atm. To adjust for the temperature, we can use the combined gas law:

V2 = (V1 × T2 × P1) / (T1 × P2)

Where:

Thus:

V2 = (22.4 L × 310 K) / 273 K

V2 is the volume of CO2 at the new conditions, which is approximately 25.5 L/mol. Since we have 0.0184 moles of HCO^{3−}, the volume of CO2 produced would be 0.0184 moles × 25.5 L/mol = 0.4692 L or 469.2 mL.

Answer:

base and it's in solution

Explanation:

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Answer:

I should use a volumetric flask.

Explanation:

If the accuracy of the concentration is important, we need to use a volumetric flask.

Answer:

The glassware appropriate for dilution preparation is a (250 mL) volumetric flask

Explanation:

We know that molarity or concentration of a solution is the number of moles per litres of solution,  the molarity and volume of the solution in question can be used to find out how much of the stock solution to be diluted using the formula;

[tex]c = \frac{n}{v}[/tex]

Once the variables are confirmed, a 250 mL solution of 1.50M HCl can be prepared by measuring determined stock solute into a 250mL volumetric flask and the making it upto the 250mL mark. (a measuring cylinder can be used to check accuracy of volume added).

Answer:

[tex]\large \boxed{\text{a) HBr; b) K; c) 0.0503 g}}[/tex]

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:   39.10    80.41               2.016  

            2K  +  2HBr ⟶ 2KBr + H₂

m/g:     5.5       4.04

a) Limiting reactant

(i) Calculate the moles of each reactant  

[tex]\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}[/tex]

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:  

The molar ratio of H₂:K is 1:2.

From HBr:  

The molar ratio of H₂:HBr is 3:2.  

[tex]\text{Moles of H}_{2} = \text{0.049 93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{2 mol HBr}} = \text{0.024 97 mol H}_{2}[/tex]

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

[tex]\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\\text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$}[/tex]

Final answer:

The true statements about particle motion and temperature are that the average kinetic energy of the particles in a substance determines its temperature and that faster moving particles generally result in a higher temperature of the substance.

Explanation:

The relationship between particle motion and temperature is based on the kinetic-molecular theory. According to this theory, temperature is proportional to the average kinetic energy of the particles, such as molecules or atoms, in a substance. Therefore:

The average kinetic energy of the particles in a substance determines the substance's temperature.

If the particles in two pure substances have the same average speed, it does not necessarily mean they have the same temperature, since heavier particles will have more kinetic energy at the same speed, leading to different temperatures.

When two substances have the same temperature, it indicates that the average kinetic energy of the particles in both substances is the same, not that all of the particles are moving at the same speed, because particles have a range of speeds.

The speed of a substance's particles can give an indication of how hot or cold that substance is. Faster moving particles generally mean a higher temperature.

As such, the true statements are 1 and 4. Understanding that temperature is not simply a measure of particle speed, but rather the average kinetic energy involved, clarifies misconceptions about the nature of thermal energy and its measurement.

Answer:

p3=0.36atm (partial pressure of NOCl)

Explanation:

2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)  Kp = 51

lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively

[tex]Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }[/tex]

[tex]Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }[/tex]

p1=0.125atm;

p2=0.165atm;

p3=?

Kp=51;

On solving;

p3=0.36atm (partial pressure of NOCl)

The reaction,

Given values,

Pressure,

Value of Kp,

Now,

→ [tex]K_p = \frac{[NOCl]^2}{[NO]^2[Cl_2]^2}[/tex]

or,

→ [tex]K_p = \frac{[P_3]^2}{[P_1]^2[P_2]}[/tex]

By substituting the values,

   [tex]51 = \frac{[P_3]^2}{[0.125]^2[0.165]}[/tex]

   [tex]P_3 = 0.36 \ atm[/tex]

Thus the response above is appropriate.  

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Answer:

The answer is flattening

Explanation:

A physical change is generally something that affects the shape of form of the matter and a chemical change results from a chemical reaction. Flames are caused by chemical reactions, as is rust, and the process of a fruit becoming ripe. Thus, the answer is “flattening”.

In the realm of chemistry, only flattening out of the listed options is considered a physical change as it alters the condition of a substance without modifying its underlying chemical structure.

In the context of Chemistry, physical changes are processes that change the form or appearance of a substance, but not its chemical composition. Of the processes you listed: burning, rusting, flattening, and ripening, flattening is an example of a physical change. For instance, if you have a piece of aluminum foil and you flatten it, it is still aluminum foil - no new substance is created. On the other hand, burning, rusting, and ripening are chemical changes because they result in new substances being formed.

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Answer:

2) HClO3 is stronger because chlorine is more electronegative than iodine.

Explanation:

The more electronegative the element is the more strong or acidic it becomes.

Chlorine being more electronegative than Iodine makes it easier for it to pull the electron of hydrogen more strongly and hence has a higher tendency to release a H+ unit. Hence that makes it stronger.

HClO3 is stronger than HIO3 because chlorine is more electronegative than iodine.

Chloric acid (HClO3) is more stronger than Iodic acid (HIO3) because chlorine is more electronegative than iodine. As we go from top to bottom in the periodic table, the atomic size increases and electronegativity decreases.

Electronegativity is the ability of an atom to attract electron pair towards itself. Chlorine atom comes on the top whereas iodine is present lower than chlorine in the periodic table so the atomic size of chlorine is smaller and higher value of electronegativity as compared to iodine so we can conclude that Chloric acid (HClO3) is more stronger than Iodic acid (HIO3).

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Answer:

No, it is not.

Explanation:

Most solutions do not behave ideally. Designating two volatile  substances as A and B, we can consider the following two cases:

Case 1: If the intermolecular forces between A and B molecules are weaker than  those between A molecules and between B molecules, then there is a greater tendency  for these molecules to leave the solution than in the case of an ideal solution. Consequently,  the vapor pressure of the solution is greater than the sum of the vapor  pressures as predicted by Raoult’s law for the same concentration. This behavior gives  rise to the positive deviation.

Case 2: If A molecules attract B molecules more strongly than they do their own  kind, the vapor pressure of the solution is less than the sum of the vapor pressures as  predicted by Raoult’s law. Here we have a negative deviation.

The benzene/toluene system is an exception, since that solution behaves ideally.

No, in the case of an ideal solution of two volatile liquids the two substances at different state cannot coexist for a single value of pressure at a given temperature.

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Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  = [tex]\frac{[SO42-] [H3O+]}{[HSO4-]}[/tex]

0.011 = [tex]\frac{x (0.1 + x)}{0.1 - x}[/tex]

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = [tex]{ -b +,-\sqrt{b^2 - 4ac[/tex] / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = [tex]\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }[/tex]  / 2(-1)

x = 0.111 +,- [tex]\sqrt{0.012321 + 0.0044}[/tex] / -2

x = 0.111 +,- [tex]\sqrt{0.016721}[/tex] / -2

x = [tex]\frac{0.111 +, - 0.1293}{-2}[/tex]

x = [tex]\frac{0.111 + 0.1293}{-2}[/tex]   , x = [tex]\frac{0.111 - 0.1293}{-2}[/tex]

x = [tex]\frac{0.2403}{-2}[/tex]    , x = [tex]\frac{0.0183}{-2}[/tex]

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

Final answer:

The pH of a 0.100 M H2SO4 solution is approximately 1.00, considering the complete ionization of the first proton and the partial ionization of the second proton, which is less significant due to its lower Ka value.

Explanation:

The question asks about the pH of a 0.100 M H2SO4 solution, taking into account the ionization of both protons. Sulfuric acid (H2SO4) is a strong diprotic acid that dissociates completely for the first proton, yielding a concentration of 0.100 M H+ and 0.100 M HSO4-. The second proton dissociation is less extensive, with a given Ka of 1.1x10-2, which we need to include in our calculation of pH.

First step ionization (complete dissociation):
H2SO4 → H+ + HSO4-

Second step ionization (partial dissociation):
HSO4- ↔ H+ + SO42- (Ka = 1.2 x 10-2)

To calculate the pH, we first consider the complete ionization of the first proton, which directly gives us 0.100 M of H+. The pH contribution from this ionization is pH = -log(0.100) = 1.00. Then we consider the second ionization of HSO4-. Given the Ka and the initial HSO4- concentration of 0.100 M, we can set up an equilibrium expression to find the additional contribution of H+ from the second ionization. However, because the ionization is low, the change in concentration of H+ due to the second ionization can be negligible for this approximate calculation. Therefore, we can assume the pH of the solution largely results from the first dissociation.

Therefore, the answer is that the pH of the 0.100 M H2SO4 solution is approximately 1.00.

Answer:

The amount that oxidized to NiO(OH) is = 1.46 gm

Explanation:

Given data

Current I = 0.35 A

Time taken = 145 min

We know that charge

Q = I t

Q = 0.35 × 145 × 60

Q = 3045 C

Faraday's constant = 96500 C

No. of  moles of electron

[tex]N = \frac{3045}{96500}[/tex]

N = 0.03155

1 mol of [tex]NI(OH)_{2}[/tex] is oxidized by 2 moles of electrons,  so no. of moles can be oxidized is

[tex]\frac{0.03155}{2}[/tex] = 0.015775 moles

Now convert this moles into gm by multiplying 92.708 [tex]\frac{gm}{mol}[/tex]

0.015775 × 92.708 = 1.46 gm

Therefore the amount that oxidized to NiO(OH) is = 1.46 gm

The given question is incomplete. The complete question is :

Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equation is balanced and contains state symbols after every reactant and product

[tex]HNO_3(aq)+H_2O(l)\rightarrow[/tex]

Answer: The complete equation is [tex]HNO_3(aq)+H_2O(l)\rightarrow NO_3^-(aq)+H_3O^+(aq)[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

[tex]HNO_3[/tex] being a strong acid dissociates to give [tex]H^+[/tex] ions an [tex]H_2O[/tex] will act as base and accept [tex]H^+[/tex] to form [tex]H_3O^+[/tex]

Thus the complete equation is [tex]HNO_3(aq)+H_2O(l)\rightarrow NO_3^-(aq)+H_3O^+(aq)[/tex]