A voltaic cell is constructed with an Ag/Ag half-cell and a Pb/Pb2 half-cell. The silver electrode is positive. Write the balanced half-reactions and the overall reaction. Include the phases of each reactant and product.

Answer:

[tex]\boxed {\boxed {\sf 8.22 \ \textdegree C}}[/tex]

Explanation:

The question asks us to calculate the temperature of the gas at the higher volume. Since the pressure is constant, we are only concerned about volume and temperature. We will use Charles's Law. This states that the volume of a gas and the temperature of the gas have a directly proportionate relationship. The formula is:

[tex]\frac {V_1}{T_1}= \frac{V_2}{T_2}[/tex]

The gas starts at a volume of 4.17 liters and a temperature of 7.62 degrees Celsius.

[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{V_2}{T_2}[/tex]

The gas is expanded to a new volume of 4.50 liters, but the temperature is unknown.

[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{4.50 \ L}{T_2}[/tex]

We want to solve for the temperature at a higher volume. We must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

[tex]4.17 \ L * T_2 = 4.50 \ L * 7.62 \textdegree C[/tex]

The variable is being multiplied by 4.17 liters. The inverse of multiplication is division. Divide both sides by 4.17 L.

[tex]\frac {4.17 \ L * T_2 }{4.17 \ L}= \frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]

[tex]T_2=\frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]

The units of liters (L) cancel.

[tex]T_2=\frac{4.50 * 7.62 \textdegree C}{4.17 }[/tex]

[tex]T_2=\frac{34.29}{4.17 } \textdegree C[/tex]

[tex]T_2=8.22302158273 \textdegree C[/tex]

The original measurements of liters and temperature have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

The 3 to the right in the thousandth place tells us to leave the 2 in the hundredth place.

[tex]T_2 \approx 8.22 \ \textdegree C[/tex]

The temperature of the gas at the higher volume is approximately 8.22 degrees Celsius.

Answer:

absorption, fluorescence = phosphorescence

Explanation:

Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.

Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.

The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.

Answer:

(d) the atoms it contains and the way these atoms are connected

Explanation:

hope it will be helpful for you

Explanation:

Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present. Physical properties can be measured without changing a substance's chemical identity.

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

Explanation:

Answer:

R.F.M of Iron (II) oxide :

[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]

Moles :

[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]

Molecules :

[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]

The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.

Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.

The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.

Given, the mass of the iron oxide = 35.2 ×10⁻²³ g

The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol

71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³

35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)

The number of molecules of FeO in a given mass = 2.95 molecules

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Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

Answer:

Explanation:

We can calculate the Rf values by using the following formula:

Rf = Distance from the baseline / Solvent front distance

With that in mind we now proceed to calculate the Rf value for both components:

Answer:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Explanation:

The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:

In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).

The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Explanation:

The term "basic unit of matter "refers to atom

A Atom

Answer:

a) Hence, T = 207 K.

b) Hence, T2 = 226 K.

Explanation:

Now the given,

n = 0.27 moles ; P = 2.5 atm ; T = 298 K

a) γ = 5/3 since Ne is a monoatomic gas.

[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]

Hence, T = 207 K

b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]

[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]

But P = P2

[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]

This gives us:

[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]

Hence, T2 = 226 K

Answer:

sosksjsjjs

Explanation:

even i know how to type şïllily

Answer:

Explanation:

a) Ionic

Lithium oxide

b) Covalent

[tex]$\ce{N_2O_3}$[/tex]

c) Covalent

Phosphorus trichloride

d) Ionic

[tex]Mn_2O_3[/tex]

Answer:

The correct answer is C. Nucleic acid

Explanation:

Nucleic acids are biological polymers which play an important role in the storage and expresion of genetic information. There are two types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are basically composed of:

- a 5-carbon sugar: deoxyribose in DNA and ribose in RNA

- phosphate group

- a nitrogenous base: adenine, cytosine, guanine and thymine in DNA; while RNA contains adenine, cytosine, guanine and uracil.

Answer:

Acids react with most metals to form hydrogen gas and salt. ... When an acid reacts with metal, salt and hydrogen gas are produced

Answer:

1) 2.69 * 10²³ PBr₅

2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁

Explanation:

Question 1)

We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.

Phosphorus pentabromide is given by PBr₅.

To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.

To convert from grams to moles, we will find the molar mass of PBr₅.

Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:

[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]

And since we want to convert from grams to moles, we can write the following ratio:

[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]

Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.

To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.

So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:

[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]

In combination, starting with 192 grams of PBr₅, we will acquire:

[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]

Cancel like units:

[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]

Multiply. Hence:

[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]

Since the final answer should have three significant digits, our final answer is:

[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]

So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.

Question 2)

We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.

3.42 kilograms is equivalent to 3420 grams of table sugar.

Again, we can convert from grams to moles, and then from moles to molecules.

First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:

[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]

To cancel units, we can write our ratio as:

[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]

With grams in the denominator.

And by definition:

[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]

Combining the two ratios and the starting value, we acquire:

[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]

Cancel like units:

[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]

Multiply:

[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]

Rewrite:

[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]

The resulting answer should have three significant digits. Hence:

[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]

So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.

Answer:

2.69×10²³ molecules of PBr₅

6.02×10²⁴ molecules of C₁₂H₂₂O₁₁

Explanation:

To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.

[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]

Now, we know that there are about 2.69×10²³ molecules of PBr₅.

To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.

[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]

Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.

Answer: The balanced equation for the complete oxidation reaction of ethanol is [tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

Explanation:

Combustion is the chemical process where an organic molecule reacts with oxygen gas present in the air to produce carbon dioxide and water molecules.

It is also known as an oxidation reaction because oxygen is getting added.

The chemical equation for the oxidation of ethanol follows:

[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

By stoichiometry of the reaction:

1 mole of ethanol reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 3 moles of water molecules.

Answer:

.87

Explanation:

p = m/V

43.5/50

.87

Answer:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Double replacement reaction.

It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Explanation:

Hello there!

In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Regards!

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

Answer:

Density of liquid = 1.59 g/cm³

Explanation:

We'll begin by converting 0.610 L to cm³. This can be obtained as follow:

1 L = 1000 cm³

Therefore,

0.610 L = 0.610 L × 1000 cm³ / 1 L

0.610 L = 610 cm³

Finally, we shall determine the density of the liquid. This can be obtained as follow:

Volume of liquid = 610 cm³

Mass of liquid = 972 g

Density of liquid =?

Density = mass / volume

Density of liquid = 972 / 610

Density of liquid = 1.59 g/cm³

The density of any given liquid is equivalent to the mass of the liquid divided by the volume of the liquid.

From the given information, we have:

The mass of the unknown liquid to be = 972 g

The volume of the unknown liquid to be = 0.610 L = 610 cm³

If the formula for calculating [tex]\mathbf{Density = \dfrac{Mass}{Volume}}[/tex]

Then;

[tex]\mathbf{Density \ of \ the \ unknown \ liquid = \dfrac{972 \\ g}{610 \ cm}}[/tex]

[tex]\mathbf{Density \ of \ the \ unknown \ liquid = 1.593442623 \ g/cm}[/tex]

The Density of the unknown liquid ≅ 1.593  g/cm³

In conclusion, the density of the unknown liquid is 1.593  g/cm³ to 3 significant figures.

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Answer:

56.8%

Explanation:

The reaction of the problem is 1:1. That means 1 mole of 2-methylcyclohexanone produce 1 mole of 2-methylcyclohexanol.

Percentage yield is defined as 100 times the ratio between actual yield of the reaction (5.14g) and the theoretical yield.

The theoretical yield (All reactant produce products) is obtained from the mass of the reactant as follows:

Theoretical Yield:

8.89g 2-methylcyclohexanone * (1mol/112.17g) = 0.07925 moles 2-methylcyclohexanone

Assuming all reactant produce the product in a 100% of yield, the moles of 2-methylcyclohexanol are 0.07925 moles and the mass (Theoretical yield) is:

0.07925 moles 2-methylcyclohexanol * (114.19g/mol) = 9.05g

Percentage yield:

5.14g / 9.05g * 100 = 56.8%

The percentage of the mass successfully converted into a new product is 57.2%.

The mass of the reactant (2-methylcyclohexanone) before the reduction is given as 8.89 g.

The mass of the product ( 2-methylcyclohexanol) produced is given as 5.14 g.

The percentage of the mass successfully converted into a new product is calculated as follows;

[tex]= \frac{5.14}{8.99} \times 100\% \\\\= 57.2 \ \%[/tex]

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In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.

Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase

Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.

The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids

Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

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Answer:

3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)

Explanation:

Let's consider the balanced chemical equation that takes place when sulfuric acid solution and solid aluminum hydroxide react to form aluminum sulfate and water. This is a neutralization equation.

3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)

Answer:

[tex]Q=292240.8J=292.2kJ[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to use the general heat equation:

[tex]Q=mC(T_2-T_1)[/tex]

For us to plug the given mass, specific heat and temperature change to obtain the required heat:

[tex]Q=120g*32.91\frac{J}{g*K} (98\°C-24\°C)\\\\Q=292240.8J=292.2kJ[/tex]

Regards!

Answer:

0.030 mole

Explanation:

Mass of crucible + lid = 23.422 g

Mass of crucible + lid + compound = 24.746 g

Mass of crucible + lid + compound - water = 24.213

Mass of water = Mass of crucible + lid + compound + heat

       = 24.746 - 24.213

                = 0.533 g

Mole of water in the hydrated compound = mass of water in the compound/molar mass of water

    = 0.533/18

         = 0.0296 mole = 0.030 mole

Answer:

1360kJ are evolved

Explanation:

When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.

To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:

Moles H2 -Molar mass: 2g/mol-

90g H2 * (1mol / 2g) = 45 moles

Moles O2 -Molar mass: 32g/mol-

90g * (1mol / 32g) = 2.81moles

For a complete reaction of 2.81 moles of O2 are needed:

2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2

As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.

As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:

2.81 moles O2 * (241.8kJ / 1/2moles O2) =

The quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is

The equation for the combustion of hydrogen is given as:

[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]

Recall that:

Since the mass of 180 g is equally shared by H₂ and O₂, then:

The number of moles of the reactant can be determined as follows:

For H₂:

[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]

no of moles = 44.6 mol

For O₂:

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]

no of moles = 2.8 mol

Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:

If 1/2 moles of O₂ produces -241.8 kJ/mol of water;

Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:

[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]

[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]

= - 1358.91 kJ

≅ - 1360 kJ

Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ

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Answer:

irreversible.

I hope this will help you