Write the formula unit equation for this reac- tion occurring in water: Ammonium fluoride and magnesium chloride are mixed to form magnesium fluoride and ammonium chloride.
1. 2 NH4F (s) + MgCl2 (aq) → MgF2 (aq) + 2 NH4Cl (s)
2. 2 NH4F (aq) + MgCl2 (aq) → MgF2 (s) + 2 NH4Cl (aq)
3. 2 NH4F (aq) + MgCl2 (aq) → MgF2 (aq) + 2 NH4Cl (aq)
4. NH4F (aq) + MgCl (aq) → MgF (s) + NH4Cl (aq)
5. 2 NH3F (aq) + MgCl2 (aq) → MgF2 (s) + 2 NH3Cl (aq)
6. 2 NH4F (aq) + MgCl2 (s) → MgF2 (s) + 2 NH4Cl (aq)

Answer: The answer is D. Molecular solids are usually excellent conductors of electric current

Explanation: Molecular solids are solids that have separate individual molecules held together by intermolecular forces (Van der Waal's force) rather than bonds.

Van der Waal's forces are the weak forces that contribute to intermolecular bonding between molecules.

Examples of a molecular solid are water ice, solid carbon dioxide and white phosphorus.

Molecular solids cannot conduct electricity because they have localized electrons (election localized within the bonds in each molecule).

Final answer:

The statement that is not applicable to molecular solids is 'd. Molecular solids are usually excellent conductors of electric current' because molecular solids are poor conductors due to the lack of ions or free electrons.

Explanation:

To answer which statement is not applicable to molecular solids, let us examine each option given:

a. The units that occupy the lattice points are molecules - This is true for molecular solids.

b. The binding forces in molecular solids are dispersion forces or dispersion forces and dipole-dipole interactions - This is also true. Dispersion forces are present in nonpolar molecular solids, while polar molecular solids exhibit dipole-dipole interactions and sometimes hydrogen bonds as well.

c. Molecular solids have relatively low melting points - This statement is correct since the intermolecular forces in molecular solids are weaker than ionic or covalent bonds, resulting in lower melting points.

d. Molecular solids are usually excellent conductors of electric current - This statement is not true. Molecular solids lack ions or free electrons, which makes them poor conductors of electricity.

e. Molecular solids are soft compared to covalent solids - This is generally true because the intermolecular bonds in molecular solids are weaker than the covalent bonds in covalent solids.

Therefore, the statement that is not applicable to molecular solids is d. Molecular solids are usually excellent conductors of electric current.

The heat of reaction, ΔH°rxn, for the production of methane, CH4, can be calculated using the enthalpy change of the reaction and the stoichiometric coefficients of the reactants and products.

The heat of reaction, ΔH°rxn, for the production of methane, CH4, can be calculated using the enthalpy change of the reaction and the stoichiometric coefficients of the reactants and products. In this case, the reaction is:

C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJ

To calculate the heat of reaction, we can multiply the enthalpy change by the stoichiometric coefficient of methane in the balanced equation:

ΔH°rxn = -393.5 kJ/mol × 1 mol = -393.5 kJ

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The heat of reaction for the production of methane (CH₄) from CO₂ and H₂O is calculated to be -890.4 kJ. Using Hess's Law and the standard enthalpies of formation, the standard enthalpy of formation for CH4 is found to be -74.7 kJ/mol.

To calculate the heat of reaction for the production of methane (CH4), we will use the given thermochemical equation:

Step-by-Step Calculation:

The equation states that 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O, releasing 890.4 kJ of energy. This reaction is exothermic, so ΔΗ is negative.

The standard enthalpy of formation [tex](\Delta H^\circ _f)[/tex] of a compound is the change in enthalpy when one mole of a substance is formed from its elements in their standard states.

We know the standard enthalpies of formation at 298 K for CO₂(g) and H₂O(l), which are:
[tex]\Delta H_f^\circ(\text{CO}_2(\text{g})) = -393.5 \, \text{kJ/mol}[/tex]
[tex]\Delta H_f^\circ(\text{H}_2\text{O}(\text{l})) = -285.8 \, \text{kJ/mol}[/tex]

Using Hess's Law: [tex]\Delta H^\circ_{\text{rxn}} = \left[ \sum \Delta H_f^\circ (\text{products}) \right] - \left[ \sum \Delta H_f^\circ (\text{reactants}) \right][/tex]

Insert values into the equation:

[tex]\Delta H^\circ_{\text{rxn}} = \left[ (-393.5 \, \text{kJ/mol}) + 2(-285.8 \, \text{kJ/mol}) \right] - \left[ \Delta H_f^\circ(\text{CH}_4(\text{g})) + 2(0 \, \text{kJ/mol for O}_2) \right][/tex]

[tex]\Delta H^\circ_{\text{rxn}} = [-393.5 - 571.6] - [\Delta H_f^\circ(\text{CH}_4(\text{g})) + 0][/tex]

[tex]\Delta H^\circ_{\text{rxn}} = -965.1 \, \text{kJ} - \Delta H_f^\circ(\text{CH}_4(\text{g}))[/tex]

Given [tex]\Delta H^\circ_{\text{rxn}}[/tex] = -890.4 kJ, solve for [tex]\Delta H_f^\circ(\text{CH}_4(\text{g})): -965.1 \, \text{kJ} - \Delta H_f^\circ(\text{CH}_4(\text{g})) = -890.4 \, \text{kJ}[/tex]

ΔHf°(CH₄(g)) = -965.1 kJ + 890.4 kJ

ΔHf°CH₄(g)) = -74.7 kJ/mol

Therefore, the standard enthalpy of formation for methane, CH₄, is -74.7 kJ/mol.

Complete Question:

You have a 275 mg sample of a compound that contains aluminum and a group 7A element (F, Cl, Br, I or At). This compound reacts with excess silver nitrate as follows (“X” represents the unknown element): AlX3 + 3 AgNO3 → 3 AgX + Al(NO3)3. This reaction forms 581 mg of AgX. Identify element X.

Answer:

Br

Explanation:

By the stoichiometry of the reaction, 1 mol of AlX3 will form 3 moles of AgX. If we call the molar mass of X as m, and by the periodic table the molar masses of Al is 26.982 g/mol, and of Ag is 107.87, thus the molar masses of the compounds will be:

AlX3 = 26.982 + 3m

AgX = 107.87 + m

And the mass relationship will be the number of moles multiplied by the molar mass. So, by a simple direct rule of three:

26.982 + 3m g of AlX3 ------------- 3*(107.87 + m) g of AgX

275 mg of AlX3             ----------- 581 mg

825*(107.87 + m) = 15676.542 + 1743m

88992.75 + 825m = 15676.542 + 1743m

-918m = -73316.208

m = 79.86 g/mol

So, comparing with the values of the masses of the group 7 elements, X must be Br (79.904 g/mol).

Answer: The mass of the new compound formed will be 159 grams.

Explanation:

Synthesis reaction is defined as the reaction in which smaller substances combine in their elemental state to form a larger substance.

[tex]A+B\rightarrow AB[/tex]

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the synthesis of copper oxide follows:

[tex]2Cu+O_2\rightarrow 2CuO[/tex]

Let the mass of new compound (copper oxide) formed be 'x' grams

We are given:

Mass of copper metal = 127 grams

Mass of oxygen gas = 32 grams

Total mass on reactant side = 127 + 32 = 159 g

Total mass on product side = x

So, by applying law of conservation of mass, we get:

x = 159 g

Hence, the mass of the new compound formed will be 159 grams.

Answer: a. Removing all traces of water to form a solid

Explanation:

Answer:

85.82%

Explanation:

Mass before recovery = 141 mg

Mass after recrystallization = 121 mg

Percent recovery of recrystallization = actual mass / mass of crude extract * 100 = 121 / 141 * 100 = 85.82%

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

                   [tex]C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)[/tex]

Initial:          24.0

At eqllm:    24-x            x              x

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}[/tex]

We are given:

[tex]K_p=0.040[/tex]

Putting values in above expression, we get:

[tex]0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1[/tex]

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

[tex]PV=nRT[/tex]          .........(1)

To calculate the mass of a substance, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]           ..........(2)

We are given:

[tex]P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

Putting values in equation 1, we get:

[tex]23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol[/tex]

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

[tex]7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g[/tex]

We are given:

[tex]P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

Putting values in equation 1, we get:

[tex]0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol[/tex]

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

[tex]0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g[/tex]

We are given:

[tex]P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

Putting values in equation 1, we get:

[tex]0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol[/tex]

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

[tex]0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g[/tex]

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

[tex]\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100[/tex]

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

[tex]\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%[/tex]

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Answer:

1.87 g of Al were reacted

Explanation:

We must apply the Ideal Gases law to solve the amount of moles of H₂ produced at STP

T = 273 K

P = 1 atm

P. V = n . R . T

1 atm . 2.34 L = n . 0.082 . 273 K

( 1 atm . 2.34 L ) / ( 0.082 . 273 K) = n

0.104 moles  = n

Now we can find the moles of Al with the reaction. As ratio is 3:2 we have to make a rule of three to determine the moles we used.

3 moles of H₂ came from 2 moles of Al

0.104 moles of H₂ would come from ( 0.104 . 2 ) /3 = 0.0693 moles of Al

Finally we can know the mass of Al we used ( moles . molar mass)

0.0693 mol . 26.98 g/mol = 1.87 g

To determine the grams of Al reacted with excess HCl, we need to calculate the number of moles of H2 gas produced at STP, then use stoichiometry to convert the moles of H2 to moles of Al, and finally convert the moles of Al to grams using its molar mass.

To determine how many grams of Al were reacted with excess HCl, we need to first calculate the number of moles of H2 gas produced at STP. From the balanced chemical equation, we can see that for every 2 moles of Al, 3 moles of H2 gas are produced. Since 1 mole of any ideal gas occupies 22.4 L at STP, we can convert the given volume of H2 gas to moles by dividing 2.34 L by 22.4 L/mol. This gives us approximately 0.1045 moles of H2 gas.

Using the stoichiometry of the reaction, we can then determine the number of moles of Al, which is the same as the number of moles of H2 gas produced. Finally, we can convert the moles of Al to grams using the molar mass of Al, which is 26.98 g/mol. Therefore, the number of grams of Al reacted would be approximately 0.1045 moles multiplied by 26.98 g/mol, resulting in approximately 2.82 grams of Al.

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Answer:

Rate law = k1 [N2O5]^2

Explanation:

Rate law only cares about reactants and rate law can only be determined experimentally or by using the coefficients of reactants in elementary reactions

If 2A---->B is an elementary reaction... rate law for this is rate=k[A]^2

So if we look at your question...2N2O5---->2N2O4 + 02 (I think you are missing O2 in question because otherwise equation is unbalanced)

Rate law = k1 [N2O5]^2

The rate law for the reaction 2N2O5(g) → 2N2O4(g) is rate = k1[N2O5]^2, indicating a second-order dependence on [N2O5]. For the reaction H2(g) + 2NO(g) → N2O(g) + H2O(g), the rate law is rate = k[NO]^2[H2], with the reaction being second order in NO and first order in H2.

The rate law for an elementary reaction can be written directly from the stoichiometry of the reaction. Considering the provided elementary reaction 2N2O5(g) → 2N2O4(g), the rate law would be rate = k1[N2O5]^2, which indicates that the reaction is second order with respect to N2O5.

In general, for an elementary reaction like H2(g) + 2NO(g) → N2O(g) + H2O(g), the rate law given is rate = k[NO]^2[H2], which means that the reaction is second order with respect to NO (order of 2) and first order with respect to H2 (order of 1), making the overall order of the reaction three.

It seems that the question is lacking hypothesized structures, which are provided in Figure A

Answer:

Option A has bond angles that have the highest strain beyond known tolerance

Explanation:

Note: Please find the complete question in the attachment

The hypothesized structures in the options were the proposed structures of benzene in the mid 19th century by Ladenburg (1869), Thiele (1899), Armstrong (1887), and Kekule (1865) respectively

Option A suggests sp3 hybridization of the carbon atom. sp3 hybridized atoms are oriented at 109.5° to be stable. The structure suggests that bond angles between C-C bonds are 60° and 90°, these bond angles are well beyond the known tolerable angle strain.

Options B and D (currently accepted structures of benzene) are sp2 hybridized and their bonds angle is 120°, which is the optimal angle for sp2 hybridized atom. Option C also suggests sp2 hybridized carbon, so their will also be no ring strain in it. But it is not an accepted structure as it was suggested to contain radical like properties by Armstrong.

The strain in molecular geometries such as 1-cyclohexyne arises from abnormal bond angles that deviate from ideal sp³ hybridization. Benzene, however, remains stable due to electron delocalization and a regular planar hexagonal structure maintaining equal bond lengths and angles.

Understanding Molecular Geometry and Strain in Hypothesized Structures

The concept of strained bond angles beyond known tolerances relates to proposed molecular structures that do not conform to normal, observed geometries. For instance, the 1,3,5-cyclohexatriene structure proposed for benzene in 1866 by Kekule features alternating single and double bonds. This would predict different bond lengths of 1.48 Å for single and 1.34 Å for double bonds. However, due to aromaticity, benzene is actually a regular planar hexagon with equal C-C bond lengths of 1.39 Å and bond angles of 120°, indicating the presence of resonance and delocalized p-orbital electrons.

In contrast, a structure like 1-cyclohexyne would suffer from severe angle strain since it attempts to maintain a planar structure with sp² hybridization that imposes 120° bond angles, which deviates from the tetrahedral angle of 109.5° expected for sp³ hybridized carbons as in cyclohexane. Additionally, cyclooctatetraene, if forced into a planar structure to achieve delocalization, would also experience prohibitive angle strain, as it can not attain a structure with equivalent π bonds between adjacent carbons without distorting natural bond angles.

The concept of electron delocalization is fundamental in understanding why certain hypothesized structures impose a strain on bond angles. For benzene, the planar structure and delocalization lead to stability, while for other cyclic compounds, the need to maintain ideal hybridization geometry can lead to significant strain if forced into non-tetrahedral geometries.

Answer: The oxidation and reduction half reactions are written below.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

For the given chemical reaction:

[tex]FeSO_4(aq.)+Zn(s)\rightarrow Fe(s)+ZnSO_4(aq.)[/tex]

The half reactions for the given reaction follows:

Oxidation half reaction:  [tex]Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-[/tex]

Reduction half reaction: [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Hence, the oxidation and reduction half reactions are written above.

The balanced half-reaction for oxidation in the reaction is Zn (s) -> Zn2+ (aq) + 2e-, signifying Zinc is being oxidised. The reduction half-reaction is Fe2+ (aq) + 2e- -> Fe (s) demonstrating Iron being reduced.

In this chemical reaction:FeSO4 (aq) + Zn (s) --> Fe (s) + ZnSO4 (aq) two processes occur simultaneously; oxidation and reduction. Oxidation involves loss of electrons, and in this case, it is the Zinc metal that gets oxidized. It unites with Sulphate forming ZnSO4 and in the process gains a positive charge by losing 2 electrons.

The balanced half-reaction for oxidation is: Zn (s) --> Zn2+ (aq) + 2e-

Reduction on the other hand involves gain of electrons. Iron in this reaction gains electrons to form metallic iron (Fe). This can be represented by the following balanced half-reaction:

Fe2+ (aq) + 2e- --> Fe (s).

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Answer: Reactants==>2 molecules of propanol.

Condition for Reaction to occur==> Strong acid,e.g HCl and high temperature (140°C).

Explanation:

Dipropyl ether can be sythesized commonly in two ways;

(1). Williamson ether synthesis of dipropyl ether: this chemical reaction involves the reaction of alkyl halide(propyl halide) with conjugate base of propanol.

(2). Acid catalyzed ether synthesis of dipropyl ether: this is the reaction this question is talking about. A strong acid is used in the Chemical Reaction, and , it is reaction between two(2) molecules of propanol. This is done in the presence of strong acid such as HCl and high temperature (up to 140°C).

The Reactants are the two molecules of propanol. The condition necessary for the reaction to occur is that the temperature must be high(140-143°C) and it must be in the presence of a strong acid.

In a dehydration reaction, 2 molecules lose one water molecule to combine.  In presence of strong acid ([tex]\bold{H_2SO4}[/tex]) and high temperature at [tex]\bold{145^oC}[/tex] propanol can be accomplished to synthesize dipropyl ether.

Dehydration of Alcohol:

In high acidic and high-temperature conditions, alcohol dehydrates to form ether.

[tex]\bold{2R-OH \rightarrow R-O-R +H_2O}[/tex]

2 molecules of propanol will combine to form dipropyl ether in the presence of strong acid ([tex]\bold{H_2SO4}[/tex]) at [tex]\bold{145^oC}[/tex].

[tex]\bold{2CH_3 CH_2CH_2OH \rightarrow CH_3 CH_2CH_2-O-CH_2CH_2CH_3 + H_2O }[/tex]

Therefore, In presence of strong acid ([tex]\bold{H_2SO4}[/tex]) and high temperature at [tex]\bold{145^oC}[/tex]

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Answer:

Pressure of alcohol after reaching equilibrium the second time = 2.62atm

Explanation:

Detailed step by step with explanation is as shown in the attachment.

Explanation:

A covalent bond is defined as the occurrence of a bond due to the sharing of electrons between the combining atoms.

Atomic number of hydrogen atom is 1 and its electronic configuration is [tex]1s^{2}[/tex]. So, in order to complete its octet it needs to gain or mutually shares one electron.

A covalent bond is generally formed between non-metal atoms.

Thus, we can conclude that hydrogen has only one electron that will be involved in the formation of a covalent bond.

In a covalent bond, a hydrogen atom contributes its single electron, which pairs with the electron from the other atom, forming a single covalent bond.

When a hydrogen atom forms a covalent bond with another atom, it shares its single electron, resulting in a paired bonding configuration. As hydrogen has one electron in its valence shell, this one electron becomes involved in bonding. When two hydrogen atoms bond, each contributes their one electron to form a molecular orbital, resulting in a single covalent bond. This is represented in the formation of a hydrogen molecule (H₂), where the bond order is calculated as (2-0)/2=1, signifying one stable covalent bond. The electron configuration for each hydrogen atom in the molecule then resembles that of helium, with two electrons completing the 1s subshell.

Answer:

0.605 molal

Explanation:

molality is the amount of solute in a particular mass of solvent.

lets calculate the amount of benzene solute.

mass of benzene= 13.3g

molar mass of C6H6= 12*6 +1*6 =72+7=78g/mol

amount of benzene= mass/molar mass

                           =13.3/78

                          =0.1705mol

molality= amount of solute/mass of solvent in kg

mass of solvent=282g=0.282kg

molality = 0.1705/0.282

    =0.605 molal

To calculate the molal concentration of benzene in the solution, divide the number of moles of benzene (0.1703 moles) by the mass of carbon tetrachloride in kilograms (0.282 kg), resulting in a concentration of 0.604 molal.

To calculate the molal concentration of benzene in the solution, first, we need to determine the number of moles of benzene. The molar mass of benzene (C6H6) is approximately 78.11 g/mol. Using the mass of benzene given (13.3 g), we can calculate the moles of benzene as follows:

Number of moles of benzene = mass of benzene / molar mass of benzene

Number of moles of benzene = 13.3 g / 78.11 g/mol = 0.1703 moles

The molal concentration is then calculated by the number of moles of solute per kilogram of solvent. Since the mass of the solvent (carbon tetrachloride) is given in grams, we convert it to kilograms:

Mass of solvent in kg = 282 g / 1000 = 0.282 kg

Now, the molal concentration (m) can be calculated as:

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

Molal concentration (m) = 0.1703 moles / 0.282 kg = 0.604 Molal

Answer:

56.2 mL

Explanation:

Given data

Assuming the gas has an ideal behavior, we can find the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1.00 atm × 250 mL/ 4.45 atm

V₂ = 56.2 mL

According to the question,

Volume,

Pressure,

By using the Boyle's Law, we get

→ [tex]P_1 V_1 = P_2 V_2[/tex]

or,

→     [tex]V_2 = \frac{P_1V_1}{P_2}[/tex]

By putting the values, we get

           [tex]= \frac{1.00\times 250}{4.45}[/tex]

           [tex]= \frac{250}{4.45}[/tex]

           [tex]= 56.2 \ mL[/tex]

Thus the above answer is right.

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Answer:

57.39 g excess Aluminum nitrite

Explanation:

When performing stoichiometric calculations, the first thing we need is the balanced chemical reaction.

In this case we will have:

Al(NO₂)₃   + 3 NH₄Cl        ⇒  AlCl₃  + 3 N₂  +  6 H₂O

( nitrite ion is NO₂⁻ )

Now that we have the balanced reaction, we need to calculate the number of moles, n, of Al(NO₂)₃   and NH₄Cl  , and perform the calculations necessary to determine the excess reagent and its amount.

The number of moles is :

n = mass / MW where MW is the molecular weight and m the mass.

MW Al(NO₂)₃ = 40.99 g/mol

MW  NH₄Cl    = 53.49 g/mol

n Al(NO₂)₃  = 102.1 g / 40.99 g/mol = 2.49 mol Al(NO₂)₃

n  NH₄Cl  = 174.3 g / 53.49 g/mol = 3.26 mol NH₄Cl  

Now lets calculate how many moles of NH₄Cl will react with 2.49 mol Al(NO₂)₃ :

( 3 mol NH₄Cl / 1 mol Al(NO₂)₃ ) x  2.49 mol Al(NO₂)₃ = 7.5 mol NH₄Cl  

We only have 3.26 mol NH₄Cl . Therefore our limiting reagent is NH₄Cl , and the excess reagent is Al(NO₂)₃

Now lets calculate the number of moles  Al(NO₂)₃  used to react with 3.26 mol NH₄Cl :

( 1 mol Al(NO₂)₃ / 3 mol NH₄Cl ) x 3.26 mol NH₄Cl =1.09 mol Al(NO₂)₃

The excess number of moles is:

= 2.49 mol - 1.09 mol =  1.40 mol  Al(NO₂)₃

grams of  Al(NO₂)₃  in excess

1.40 mol Al(NO₂)₃ x 40.99 g/mol = 57.39 g

Answer:

The spore coat (proteinous coat)

Explanation:

Endospore is usually a position taken by a gram positive bacteria when it lacks nutrient. Endo- means within and the-spore means offspring. This is a complex stage which is a dormant and highly resistive cell wall that allows the bacteria preserve its genetic materials in times of deprivation within this protective cell.

They are also resistant to UV radiation, heat and chemicals. Once the environment become favourable, they reactivate from this vegetative state and absorb nutrients. This vegetative state consists of the following layers:

1. The outer coating called the exosporium which surrounds the spore is the thin layer responsible for covering the spore coat.

2. The proteinous coating which is the spore coat is the next layer which as a sieve. It is resistant to chemical and other toxins.

3. After this proteinous coating is a thick peptidoglycan called the cortex.

3. A cell/core wall which resides under the cortex. This layer will become the cell wall of the bacterium after the endospore germinates.

5. The center/core of the endospore js the next. The core has the spore chromosomal DNA and some cell structures, such as ribosomes and enzymes but is inactive.

Answer: The quantity of every prefix is written below as a power of ten.

Explanation:

In the metric system of measurement, the name of multiples and subdivision of any unit is done by combining the name of the unit with the prefixes.

For Example: deka, hecto and kilo means 10, 100 and 1000 respectively. Deci, centi and milli means one-tenth, one-hundredth, and one-thousandth respectively.

The quantity of these prefixes are written as the power of 10.

For the given prefixes:

Nano:  The quantity will be [tex]10^{-9}[/tex]

Kilo:  The quantity will be [tex]10^3[/tex]

Centi:  The quantity will be [tex]10^{-2}[/tex]

Micro:  The quantity will be [tex]10^{-6}[/tex]

Milli:  The quantity will be [tex]10^{-3}[/tex]

Mega:  The quantity will be [tex]10^6[/tex]

Hence, the quantity of every prefix is written above as a power of ten.

Answer:

5.242

Explanation:

Molar mass of Ca = 40.078 g/mol

Molar mass of S = 32.065 g/mol

Equation of reaction

Ca (s) + S(s) - CaS(s)

To find the limiting reagent

Mole of Ca = 8.54/40.078 (g/mol) = 0.213 mol

Mole of S = 2.33g / 32.065 (g/mol) = 0.072664 mol

From their mole ratio, sulphur is the limiting reagent

From the reaction

1 mole of calcium react with 1 mole of Sulphur to yield 1 mole of calcium sulphide

0.072664 mol of calcium will react with 0.072664 of sulphur to yield 0.072664 mole of CaS

Theoretical yield of CaS = 0.072664 × molar mass of CaS = 0.072664 × 72.143 g/mol = 5.242 g

Explanation:

While working in a laboratory it is necessary to be free from any kind of distraction. This is because a distraction while working in a laboratory can lead to serious consequences.

For this it is important to put all your electronic devices like mobile phones, tablets etc in your backpack. Also, you should not carry any kind of eatable, water bottle or anything into the lab.

This is because chemicals do react with food items. Hence, then food does not remain fit for consumption.

Thus, we can conclude that pen or pencil and lab manual personal items are permitted to be at your workstation in the organic laboratory.

Complete question

A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 X 10²⁷ g. Use this information to answer the questions below. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of bricks? Round your answer to 2 significant digits. How many moles of bricks have a mass equal to the mass of the Earth? Round your answer to 2 significant digits.

Answer:

The mass of 1 mole of bricks = 24 X 10²⁶ g (2 significant digits), and

2.5 moles of bricks (2 significant digits) have a mass equal to the mass of the Earth.

Explanation:

Given mass of bricks = 4.0 kg

4.0kg of brick = 4000g

1 mole of any substance always contain  6.022 X 10²³ Avogadro's number

1 mole of bricks contains 6.022 X 10²³ bricks

Therefore, mass of 1 mole of brick = 6.022 X 10²³ X 4000g

= 4000 X 6.022 X 10²³ g

= 24088 X 10²³ g

The mass of 1 mole of bricks = 24 X 10²⁶ g (2 significant digits)

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⇒How many moles of bricks have a mass equal to the mass of the Earth?

Given mass of Earth = 6.0 X 10²⁷ g

24088 X 10²³ g --------> mass of 1 mole of bricks

6.0 X 10²⁷ g --------------> ?

= (6.0 X 10²⁷ g)/(24088 X 10²³ g)

= 2.4909 moles of bricks

≅ 2.5 moles of bricks ( 2 significant digits)

Therefore, 2.5 moles of bricks have a mass equal to the mass of the Earth

Answer

2.4x10^27 g

2.5 mol

The work done on the gas mixture when it is compressed from 99.0 L to 98.0 L at a constant pressure of 69.0 atm is -69.0 atm L.

In order to calculate the work done on a gas, we can use the equation:

Work = Pressure * Change in Volume

In this case, the pressure is held constant at 69.0 atm, and the change in volume is from 99.0 L to 98.0 L.

Using the formula, we can calculate the work as follows:

The work done on the gas mixture is -69.0 atm L. The negative sign indicates that work is done on the system rather than by the system.

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To estimate the reaction enthalpy (ΔH) for the given reaction, the energies required to break the reactant bonds are added, and the energies released from forming product bonds are subtracted. The estimated ΔH is found to be 1902 kJ/mol, which is the energy absorbed by the reaction.

The student was asked to estimate a balanced reaction using average bond energies. To solve for the reaction enthalpy (ΔH), the energy required to break the bonds in the reactants must be calculated and then the energy released in forming the bonds of the products has to be subtracted from it.

For the given reaction:

To calculate ΔH:

ΔH = Energy required to break bonds - Energy released in forming bonds

ΔH = (348 + (5 x 414) + 242) - (327 + 431)

ΔH = (348 + 2070 + 242) - 758

ΔH = 2660 - 758

ΔH = 1902 kJ/mol

Thus, the estimated ΔH for the reaction is 1902 kJ/mol, which indicates that it would absorb this amount of energy.

Explanation:

The minimum uncertainty position is 100 nm.

Therefore, by Heisenberg Uncertainty principal

ΔxΔp≥h/2

Δp≥h/(2Δx)

we know that

h= [tex]1.0546\times10^{-34}[/tex] Js

Δx= [tex]100\times10^{-12} m[/tex]

therefore,

[tex]\Delta p =\frac{1.0546\times10^{-34}}{2\times12\times10^{-12}} = 5.3\times10^{-25} kgms^{-1}[/tex]

therefore,

[tex]\Delta v= \frac{\Delta p}{m} =\frac{5.3\times10^{-25}}{9.11\times10^{-31}}= 5.8\times10^{5} ms^{-1}[/tex]

corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.

Answer:

5.37g

Explanation:

0.21M means ; 0.21mol/dm³

1dm³=1L , so we can say 0.21mol/L

if 0.21mol of Zinc nitrate is contained in 1L of water

x   will be contained in 135mL of water

x= 0.21*135*10³/1

=0.02835moles

number of moles=  mass/ molar mass

mass= number of moles *molar mas

molar mass of Zn(NO₃)₂=189.36 g/mol

mass= 0.02835 *189.36

mass=5.37g

Answer:

Explanation:

In gel filtration chromatography a porous matrix is used as the stationary phase. The smaller molecules are able to diffuse into this matrix while the larger ones are not. For this reason the larger molecules will elute first, followed by the smaller ones in decreasing order of size.

In other words the largest molecule will elute first and the smallest molecule will elute last:

Answer: 6 valence electrons

Explanation: the valence electron is the number of electron in the outermost shell of an atom.

Oxygen has an atomic number of 8. The electronic configuration will be 1s2 2s2 2p4 or 2, 6. Meaning the last shell is having 6 electrons and this is equivalent to the valence electron

Oxygen has 6 valence electrons.

Oxygen has an atomic number of 8, which tells us that it has 8 protons in its nucleus. The atomic mass of oxygen is 16, which is the weighted average mass of all the isotopes of oxygen. Since the atomic number represents the number of protons and the number of electrons in an element, oxygen also has 8 electrons.

In the electron configuration of oxygen, the electrons are arranged in energy levels or shells. The valence electrons are the electrons in the outermost energy level, also known as the valence shell. Oxygen's electron configuration is 1s2 2s2 2p4. From this configuration, we can determine that oxygen has 6 valence electrons.

Valence electrons are responsible for the chemical behavior of an atom. In the case of oxygen, its 6 valence electrons allow it to form two covalent bonds. This enables oxygen to react with other elements to form compounds such as water and carbon dioxide.

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The question is incomplete, here is the complete question:

What is the mass % of ammonium chloride in a 1.73 M ammonium chloride aqueous solution at 20 °C? The density of the solution is 1.0257 g/mL

Answer: The mass percent of ammonium chloride in solution is 9.03 %

Explanation:

We are given:

Molarity of ammonium chloride solution = 1.73 M

This means that 1.73 moles of ammonium chloride is present in 1 L or 1000 mL of solution.

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1.0257 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

[tex]1.0257g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.0257g/mL\times 1000mL)=1025.7g[/tex]

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of ammonium chloride = 1.73 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in above equation, we get:

[tex]1.73mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.73mol\times 53.5g/mol)=92.6g[/tex]

[tex]\text{Mass percent of ammonium chloride}=\frac{\text{Mass of ammonium chloride}}{\text{Mass of solution}}\times 100[/tex]

Mass of solution = 1025.7 g

Mass of ammonium chloride = 92.6 g

Putting values in above equation, we get:

[tex]\text{Mass percent of ammonium chloride}=\frac{92.6g}{1025.7g}\times 100=9.03\%[/tex]

Hence, the mass percent of ammonium chloride in solution is 9.03 %