When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from oC to 87°C

Final answer:

The concentration of the HBr solution is 0.001888 M.

Explanation:

To determine the concentration of the HBr solution, we need to use the equation:

HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l)

From the balanced equation, we can see that the mole ratio of HBr to NaOH is 1:1. Since the volume of NaOH required to reach the equivalence point is 18.88 mL, we can calculate the number of moles of NaOH used:

Moles of NaOH = concentration of NaOH (M) × volume of NaOH (L)

Moles of NaOH= 0.100 M × 0.01888 L = 0.001888 mol of NaOH

Since HBr and NaOH have a 1:1 mole ratio, the concentration of the HBr solution is also 0.001888 M.

3 is the number of water molecules associated with each [tex]Al_2(SO_4)_3[/tex] unit.

Molecules are made up of one or more atoms.

The molecular mass of Al₂(SO₄)₃·xH₂O

(27×2)+(32×3)+(16×12)+(x×18)

= 342 + 18x g

Molecular mass of Al₂:

27×2 = 54 g

54g contains 13.63% Al by mass.

(342+18x)g contains 100%

So,

0.1363 (342+18x) = 54

46.6146 + 2.4534x = 54

2.4534x = 7.3854

x ≈ 3

Hence, 3 is the number of water molecules associated with each [tex]Al_2(SO_4)_3[/tex] unit.

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Answer: The correct statements are the concentration of one or more of the products is small, the reaction will not proceed very far to the right and the reaction will generally form more reactants than products.

Explanation:

[tex]K_{eq}[/tex] is defined as the equilibrium constant of the reaction. It is basically the ratio of concentration of products to the concentration of reactants, each raised to the power their stoichiometric coefficients.

For a reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{eq}[/tex] is:

[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

When [tex]K>1[/tex], forward reaction is favored and when [tex]K<1[/tex], backward reaction is favored.

When K < 1, the expected possibilities are:

Hence, the correct statements are the concentration of one or more of the products is small, the reaction will not proceed very far to the right and the reaction will generally form more reactants than products.

Answer:

The percent of oxalic acid in a solid is 48.87%.

Explanation:

Mass of the solid sample = 0.7984 g

[tex]H_2C_4O_4+2NaOH\rightarrow Na_2C_2O_4+2H_2O[/tex]

Volume of NaOH solution = 37.98 mL = 0.03798 L

Concentration or molarity of the NaOH solution = 0.2283 M

Moles of NaOH :

[tex]Molarity\times \text{Volume of the solution} = 0.0086708 moles[/tex]

According to reaction,  2 moles of sodium hydroxide reacts with 1 mole of oxalic acid .

Then 0.0086708 moles of sodium hydroxide will react with:

[tex]\frac{1}{2}\times 0.0086708 moles=0.0043354 moles[/tex] of oxalic acid.

Mass of oxalic acid neutralized = 0.0043354 moles × 90 g/mol =0.390186 g

Percentage of oxalic acid in solid sample :

[tex]\%=\frac{\text{Mass of oxalic acid}}{\text{Mass of sample}}\times 100[/tex]

[tex]\%=\frac{0.390186 g}{0.7984 g}\times 100=48.87 \%[/tex]

The percent of oxalic acid in a solid is 48.87%.

To find the final volume of a gas when the temperature increases at constant pressure, apply Charles's Law. The final volume of the 500 ml gas sample that is heated from 150 K to 350 K is calculated to be 1166.67 ml.

The problem relates to Charles's Law, which states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its temperature in Kelvin. If we know the initial volume and temperature of a gas sample and the temperature changes, we can use Charles's Law (V1/T1 = V2/T2) to find the final volume of the gas.

To calculate the final volume of the 500 ml sample of gas that increases in temperature from 150 K to 350 K while keeping the pressure constant, we use the formula:

V1/T1 = V2/T2

Substituting in the given values:

500 ml / 150 K = V2 / 350 K

We solve for V2:

V2 = 500 ml × (350 K / 150 K)

V2 = 500 ml × (7/3)

V2 = 1166.67 ml

The final volume of the gas will be 1166.67 ml.

Sugar is a pure substance, specifically, it is classified as a compound composed of carbon, hydrogen, and oxygen atoms. As such, the answer to the multiple-choice question is (a) pure substance compound.

Sugar is a type of substance that is considered to be a pure substance compound. A compound is a chemical substance composed of many identical molecules composed of atoms from more than one element. In the case of sugar, specifically sucrose, it is composed of carbon, hydrogen, and oxygen atoms.

A pure substance is a material that has a constant composition and cannot be separated into its constituent parts without undergoing a chemical reaction. Sugar meets this definition as it is composed of the same type of molecule throughout the entire substance.

Therefore, the answer is (a) pure substance-compound.

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The enthalpy change for the combustion of ethanol can be calculated using the enthalpies of formation and applying Hess's law. It's determined by the difference in the products and reactants enthalpy of formation. The enthalpy change for the combustion of ethanol is -1366 kJ per mole of ethanol burned.

To determine the enthalpy change for the combustion of ethanol C₂H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g), we need to apply the concept known as Hess's law. In simple terms, Hess's law states that the total enthalpy change of a chemical reaction is the sum of the enthalpy changes for the steps of the process.

Here, we know the enthalpies of formation for the reactants and products involved. Enthalpies of formation are defined for one mole of a substance formed from its elements in their standard states. The enthalpies of formation for the substances involved are ethanol (C₂H5OH(l)) -278 kJ/mol, water (H₂O(l)) -286 kJ/mol, and carbon dioxide (CO2(g)) -394 kJ/mol.

The enthalpy change of the reaction ΔH = Σ[ (products moles x products enthalpy of formation) - (reactants moles x reactants enthalpy of formation) ]

On substituting the numbers we get ΔH = [ (2 mol CO₂ x -394 kJ/mol CO₂) + (3 mol H2O x -286 kJ/mol H2O) ] - [ (1 mol C2H5OH x -278 kJ/mol C2H5OH) + (3 mol O2 x 0 kJ/mol O2) ].

So, ΔH = [ -788 kJ + -858 kJ ] - (-278 kJ) = -1366 kJ

Thus, the enthalpy change for the combustion of ethanol is -1366 kJ per mole of ethanol burned.

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Answer:

The correct answer is option B, ammonia

Explanation:

A weak base solution like pyridine (C5H5N) can be neutralized in the presence of an acid only. A weak base in no case shall be neutralized by another base. Since ammonia is also a base thus it cannot neutralize another base. In a neutralized mixture, slats are produced by equal contribution from both the acid and base. However, when a weak base is placed into another base their will be more OH- ions but very rare H+ ion. Thus mixture with excess of OH- will again be a base only.

Answer : Option A) [tex] NH_{4}^{+} and NH_{3} [/tex]

Explanation : [tex] NH_{4}^{+} and NH_{3} [/tex] are considered as an acid-conjugate base pair, because[tex] NH_{4}^{+}[/tex] is an acid with a proton extra in its compound which makes it more acidic and rich in proton, while [tex] NH_{3} [/tex] is found to be the conjugate base of [tex] NH_{4}^{+} [/tex] as it does not contains the [tex]H^{+}[/tex] ions.

[tex] NH_{4}^{+} and H_{3}O^{+} [/tex] both are acids, and the last option where both [tex] H_{2}O and NH_{3} [/tex] can be considered as bases.

The reaction which is an acid-conjugate base pair is NH4+. This is the conjugate acid of NH3

An acid-conjugate base pair is a pair of substances that would have been similar in chemical composition but for the presence of one proton (H+).

To form a conjugate Acid, one proton must be added to a base. To create a conjugate Base on the other hand, one proton must be removed from an acid.

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There are 1.08 grams of co2 contained in 550 ml of the gas at stp

Given that    Volume of gas   =550 ml   = 0. 550  L

                 Standard Temperature  =  T  = 273.15 K

                 Standard Pressure  =  P  =  1 atm

                 Gas Constant  =  R  =  0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹

        Mass=?

From the question,    the CO₂ gas is acting ideally so we will apply Ideal Gas equation;

                              P V  =  n R T

Solving for n,

                             n  =  P V / R T

Putting values,

               n  =  1 atm × 0.55 L / 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15 K

               n  =  0.0245 moles

Remember that  

Number of Moles  =  Mass / Molar.Mass

Solving for Mass,

Mass  =  Moles × M.Mass

Mass  =0.0245 mol × 44.01 g/mol

Mass = 1.08 grams of CO₂

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Answer: The molecular formula for this compound is [tex]Na_3PO_4[/tex]

Explanation:

We are given a chemical compound known as sodium phosphate. It is formed by the interaction of oppositely charged ions.

The ions forming this compound are sodium ion [tex](Na^+)[/tex] and phosphate ion [tex](PO_4^{3-})[/tex]

To know the chemical formula for a compound, we use criss-cross method, in which the charge of the ion becomes the subscript of the other ion.

Hence, the molecular formula for given compound will be [tex]Na_3PO_4[/tex]

Answer:

Enthalpy of combustion (for 1 mol of butane)=-2657.4 [tex]\frac{kJ}{mol}[/tex]

Explanation:

Combustion is a rapid oxidation chemical process that is accompanied by low energy shedding in the form of heat and light. Oxygen is the essential element for oxidation to occur and is known as a oxidizer. The material that oxidizes and burns is the fuel, and is generally a hydrocarbon, as in this case butane C4H10 (g)

The balanced reaction is:

2 C4H10   +      13 O2       →      8 CO2   +     10H2O

Note that a balanced equation must have the same amount of each atom in the reagents and in the products, as in the previous reaction.

The heat of formation is the increase in enthalpy that occurs in the formation reaction of one mole of a certain compound from the elements in the normal physical state (under standard conditions: at 1 atmosphere of pressure and at 25 degrees of temperature).

In literature you can obtain the following heats of formation of each of the molecules involved in the reaction:

Heat of formation of C4H10 = -125.7 kJ/mol

Heat of formation of water = -241.82 kJ/mol

Heat of formation of CO2 = -393.5 kJ/mol

For the formation of one mole of a pure element the heat of formation is 0, in this case we have as a pure compound the oxygen O2

You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. For that you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

                                             = (-393.5X8) + (-241.82X10) - (-125.7X2)

                                             = -5314.8 kJ/mol

But, if you observe the previous balanced reaction, you can see that 2 moles of butane are necessary in combustion. And the calculation of the heat of reaction previously carried out is based on this reaction. This ultimately means that the energy that would result in the combustion of 2 moles of butane is -5314.8 kJ/mol.

Then, applying a rule of three can calculate energy required for the combustion of one mole of butane: if for the combustion of two moles of butane an enthalpy of -5314.8 kJ / mol is required, how much energy is required for the combustion of one mole of butane?

Enthalpy of combustion (for 1 mol of butane)=[tex]\frac{-5314.8 \frac{kJ}{mol} }{2}[/tex]

Enthalpy of combustion (for 1 mol of butane)=-2657.4 [tex]\frac{kJ}{mol}[/tex]

The change in reaction rate when doubling the concentration of [b] is dependent on the rate law of the reaction. For one-to-one stoichiometry, the reaction rate should double. However, reaction orders for more complex reactions may alter the rate differently.

In changing the concentration of reactant [b] in any reaction, the change in reaction rate is dependent on what is known as the 'rate law,' and cannot be predicted without knowledge of this law. In simple reactions, typically found in elementary chemistry, if the reaction has a one-to-one stoichiometry, like aA → bB, doubling the concentration of a reactant, in this case [b], will double the rate of reaction. This means, if changes in concentration of [b] only affect the reaction rate, then doubling [b] should double the reaction rate. However, for more complex reactions, we use the concept of reaction orders. The exponent on the concentration term in the rate law, sometimes referred to as 'n' in this context, dictates how a change in [b] alters the reaction rate. If n=1, doubling [b] would indeed double the reaction rate, but if n=2, doubling [b] would quadruple the rate, and so forth.

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The reaction rate of the provided rate law changes by a factor of the square root of two, or approximately 1.41, when the concentration of reactant B is doubled, all else remaining constant.

It refers to the change in reaction rate given a rate equation when the concentration of one reactant (B) is doubled, with the other reactant concentrations held constant. According to the provided rate law,

Rate=k [B][tex]^{\frac{1}{2}}[/tex] [C]²,

when the concentration of B is doubled, the reaction rate changes by the square root of two. Since the order of reaction concerning B is 1/2, the effect of doubling the concentration of B can be represented mathematically as:

New rate when [B] is doubled:

[tex]k(2[B])^{\frac{1}{2}} [C]^{2}[/tex] = 2[tex]^{ \frac{1}{2} }[/tex]R

The factor changes in rate = 1.41R

Therefore, the reaction rate changes by a factor of [tex]\sqrt{2}[/tex] or approximately 1.41, when the concentration of B is doubled.

The right answer is:

*Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an N, O, or F atom.

*A hydrogen bond is possible with only certain hydrogen-containing compounds.

*A hydrogen atom acquires a partial positive charge when it is covalently bonded to an F atom.

Watch the animation and identify the correct conditions for forming a hydrogen bond. check all that apply..

This animation showing the transient hydrogen bonding between water molecules.Each water molecule (H2O) has two hydrogens and two lone pairs of electrons on the oxygen, therefore it can make up to four hydrogen bonds with neighbouring molecules.

Grade:  9

Subject:  Chemistry

Chapter:  hydrogen bond

Keywords: hydrogen bond, covalent bond, hydrogen-containing compounds, partial positive charge, compounds

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an n, o, or f atom.

A hydrogen bond is possible with only certain hydrogen-containing compounds.

A hydrogen atom acquires a partial positive charge when it is covalently bonded to an f atom.

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an nitrogen, oxygen, or fluorine atom in which hydrogen shares its electron with these atoms and as a result both atoms get stability. In this type bonding, no complete transfer of electron occurs.

But the electron moves to the atom that have the highest electronegativity value which results in bearing of partial positive charge on hydrogen and partial negative charge on the atom which attracts that electron so we can conclude that hydrogen atom acquires a partial positive charge when it is covalently bonded to nitrogen, oxygen, or fluorine atom atom.

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Kepler discovered the laws of planetary motion, including orbits being ellipses, equal areas swept out in equal times, and the period of revolution being related to the semi-major axis.

Kepler discovered that:

Each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus.

The radius vector from Sun to planet sweeps out equal areas in equal times.

The planet's period of revolution squared is proportional to the cube of the semi-major axis of the ellipse.