The average C - H bond energy in CH4 is 415 kJ / mol. Use the following data to calculate the average C - H bond energy in ethane ( C2H6;C - C bond ) , in ethane ( C2H4;C C bond ) , and in ethyne ( C2H2;C C bond ) .
C2H6 ( g ) + H2 ( g ) rightarrow 2CH4 ( g ) ΔHrxn = - 65.07 kj / mol
C2H4 ( g ) + H2 ( g ) rightarrow 2CH4 ( g ) ΔH = - 202.21kJ / mol
C2H2 ( g ) + 3H2 ( g ) rightarrow 2 CH4 ( g ) ΔH = - 376.74kj / mol
Average C - H bond
Average C - H bond energy in ethane = kJ / mol energy in ethane = kj / mol
Average C - H bond energy in ethyne = kJ / mol

Answer:

230

Explanation:

The company declared 15% stock dividend, each share holder will receive an increment of 15 shares per 100 shares, since Hona owns 200 shares she will receive 30 more shares so that her total share will 230

Final answer:

The final temperature of the mixture will be 15.44°C.

Explanation:

In this problem, we can use the principle of conservation of energy to find the final temperature of the mixture. The heat gained by the ice cubes is equal to the heat lost by the water.

To find the heat gained by the ice cubes, we can use the formula:

Q = m × c × ΔT

where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Since the ice cubes are at 0°C, the heat gained is:

Q = 4 × (53.5 g) × (4.184 J/(g°C)) × (0°C - T-f)

where T-f is the final temperature.

To find the heat lost by the water, we can use the formula:

Q = m × c × ΔT

where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Since the water is at 75°C, the heat lost is:

Q = (115 g) × (4.184 J/(g°C)) × (T-f - 75°C)

Since no heat is lost to the surroundings, the heat gained by the ice cubes is equal to the heat lost by the water. Setting the two equations equal to each other, we can solve for T-f:

4 × (53.5 g) × (4.184 J/(g°C)) × (0°C - T-f) = (115 g) × (4.184 J/(g°C)) × (T-f - 75°C)

Simplifying the equation gives:

14.0816 × T-f = 496.82 - 18.0714 × T-f

32.1526 × T-f = 496.82

T-f = 15.44°C

Answer:

1. Inversely proportional

2. Option C. Boyle's Law

3. Directly proportional

4. Option C. Gay-Lussac's Law

5. Directly proportional

6. Option C. Charles' Law

Explanation

Boyle's law states that the volume of a fixed mass of gas is inversely proportional to the pressure provided temperature remains constant. Mathematically,

V & 1/P

V = K/P

PV = K(constant)

Charles' law states that the volume of a fixed mass of gas is directly proportional to it's absolute temperature, provided pressure remains constant. Mathematically,

V & T

V = KT

V / T = K(constant)

Gay-Lussac's Law states that the pressure of a fixed mass of gas is directly proportional to it's absolute temperature, provided the volume remains constant. Mathematically

P & T

P = KT

P/ T = K (constant)

Answer:

formula for NO2 is nitrogen dioxide

Answer:

Critical Point

Explanation:

You can no longer distinguish a liquid from a gas when a object hits it's critical point.

Answer:

9 atm is the total pressure of the combined gases.

Explanation:

According to the Dalton's law of partial pressure , the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

[tex]P_T=p_{1}+p_{2}....p_{n}[/tex]

where,

[tex]P_T[/tex] = total pressure =

[tex]p_{1}[/tex] = partial pressure of gas-1

[tex]p_{2}[/tex] = partial pressure of gas -2

[tex]p_{n}[/tex] = partial pressure of nth gas

We have :

Pressure of the oxygen gas in flask before mixing = 8 atm

Pressure of the hydrogen gas in flask before mixing = 1 atm

Partial pressure of oxygen gas after mixing = [tex]p_1=8 atm[/tex]

Partial pressure of hydrogen gas after mixing = [tex]p_2=1 atm[/tex]

Total pressure of the mixture : P

[tex]P=P_1+P_2[/tex] (Dalton's law of partial pressure)

[tex]P=8 atm+1 atm=9 atm[/tex]

9 atm is the total pressure of the combined gases.

Answer:

D

Explanation:

Answer:

Molecular formula of the compound is C₃H₆O

Explanation:

Firstly let's determine the moles of the vapor (gas) with the Ideal Gases Law, so it can give us the molar mass with the mass and afterwards we can work with the percent composition.

Pressure . Volume = moles . Ideal Constant Gases . Temperature in K

Temperature in K = T°C + 273 → 150°C + 273 = 423K

P . V = n . R . T

n = (P .V) / (R. T)

n = 1 atm . 0.5L / (0.082 . 423K)

n =  0.0144 moles

These are the moles for 0.8365 g, so let's determine the molar mass

Molar mass (g/mol) = 0.8365 g / 0.0144 mol → 58.02 g/mol

Percent composition means:

100 g of compound have 62.04 g of C

100 g of compound have 10.41 g of H

100 g of compound have 27.54 g of O

Let's make the rule of three:

100 g of compound have __ 62.04 g of C __ 10.41 g of H __ 27.54 g of O

The 58.02 g of compound must have:

(58.02 g . 62.04 g) / 100 g  = 36 g of C

(58.02 g . 10.41 g) / 100 g  = 6 g of H

(58.02 g . 27.54 g) / 100 g  = 16 g of O

Let's find out the moles of each

Mass / Molar mass

36 g / 12 g/mol = 3 C

6 g / 1 g/mol = 6 H

16g / 16 g/mol = 1 O

The molecular formula of the given compound is C₃H₆O. The molecular formula can be determined by finding the ratio of each element in the compound.

It can be determined by finding the ratio of each element in the compound.

First, calculate the moles of the compound from the ideal gas formula,

[tex]n = \dfrac {1 {\rm\ atm \times 0.5L} }{(0.082 \times 423{\rm \ K})}\\\\n = 0.0144 \rm \ moles[/tex]

Then calculate the molar mass of the compound,

[tex]m = {\rm \dfrac {0.8365 \ g }{0.0144 \ mol}} \\\\m = 58.02 \rm \ g/mol[/tex]

Then calculate the mass of individual elements in the 58.02 g of compound:

[tex]\text{ Mass of Carbon} = \dfrac {58.02 {\rm \ g} \times 62.04 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Carbon} = 36 \rm \ g[/tex]

[tex]\text{ Mass of Hydrogen } = \dfrac {58.02 {\rm \ g} \times 10.14 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Hydrogen } = 6 \rm \ g[/tex]

[tex]\text{ Mass of Oxygen } = \dfrac {58.02 {\rm \ g} \times 27.54 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Oxygen } = 16 \rm \ g[/tex]

Now find the moles of each element, we get

3 moles of Carbon

6 moles of Hydrogen

1 mole of Oxygen

Therefore, the molecular formula of the given compound is C₃H₆O.

Learn more about the molecular formula,

https://brainly.com/question/14666499

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

Final answer:

To find the empirical formula of a compound from its combustion products, convert the masses of carbon dioxide and water to moles to determine the moles of carbon, hydrogen, and oxygen in the compound. Calculating these mole ratios leads to determining that the empirical formula of the compound is C3H6O.

Explanation:

To determine the empirical formula of the compound given its combustion products, begin by converting the mass of carbon dioxide (CO2) and water (H2O) to moles. This reveals the moles of carbon and hydrogen in the original compound. Since oxygen is also part of the compound, calculate its moles by subtraction from the total mass of the original compound.

Convert 52.8 g of CO2 to moles: (52.8 g) / (44.01 g/mol) = 1.2 mol of C.Convert 21.6 g of H2O to moles: (21.6 g) / (18.015 g/mol) = 1.2 mol of H2, or 2.4 mol of H.Calculate moles of oxygen in the compound: Subtract the mass of C and H in the original compound from its total mass. Mass of C from CO2 = 1.2 mol × 12 g/mol = 14.4 g; Mass of H from H2O = 2.4 mol × 1 g/mol = 2.4 g. Total mass of C and H = 14.4 g + 2.4 g = 16.8 g; Mass of O = 23.2 g (total mass) - 16.8 g = 6.4 g, which is (6.4 g) / (16 g/mol) = 0.4 mol of O.To find the empirical formula, divide each mole value by the smallest number of moles: C=1.2/0.4, H=2.4/0.4, O=0.4/0.4, giving a ratio of C: 3 H: 6 O: 1. Therefore, the empirical formula is C3H6O.

Answer:

The answer to this question is 3.  Kinetic energy decreases as temperature decreases.

Explanation:

The average Kinetic energy is given by

[tex]K = \frac{3}{2}[/tex] × [tex]K_{B}[/tex] × T where

K = The average molecular kinetic energy of the gas (J)

[tex]K_{B}[/tex] = Boltzmann's constant (1.38×[tex]10^{-23}[/tex] J/K)

T = Temperature of the gas in Kelvin (k)

When the temperature of a given mass of gas drops average kinetic energy of the molecules goes down and the average molecular speed decreases.  The lower kinetic energy  which is the energy of motion of the molecules indicates lower speed of the molecules.

Note that, the kinetic energy and average speed are related by the following formula

KE = [tex]\frac{1}{2}[/tex]×m×v²

Where KE = Kinetic energy in Joules (J)

v = the velocity in m/s and

m = the mass of the molecule in Kg

Answer:

Answer is C

Explanation:

Edge 2020

Answer:

calculate the molecular formula of a compound with the empirical formula CH2O and a molar mass of 150g/mol

the molecular formula is [tex]C_{5} H_{10}O_{5}[/tex]

The molecular formula of a compound is the formula comprising of the constituent elements chemical symbols each of which carries the number of atoms of that element present in a molecule of the compound appearing in the smallest whole number ratio to other eatoms present in the compound

Explanation:

The masses of the constitent element is determined forst from which the number of moles is then calculated by dividing the mass by the molar mass then then each calculated molar mass value is divided by the smallest  number of moles calculated from the previous step so the value calculated is then rounded up to the nearest whole number giving the ratios of the moles of the elements in the compound which represents the subscripts in the empirical formula of the compound.

If the subscrips are in fractions, then multiply each of them by the same number to derive the smallest whole number factor, that is if the calculated formula contains a facor of 0.5, multiply by 2

Mass pf Carbon  = 12g

mass of Hydrogen = 1g

molar mass of oxygen - 16g

Total mass of CH2O = 30g

Therefore dividing molar mass by empirical formula mass = 150g/30g = 5

Hence our factor is 5

multiplying each subscript of the empirical formula by 5 gives

C5H10O5 hence the molecular formula is [tex]C_{5} H_{10}O_{5}[/tex]

To calculate the rate constant (k) at 75°C, we need the frequency factor (A) for the reaction, which is not provided in the question.

The rate constant (k) of a reaction can be calculated using the Arrhenius equation:

k = Ae-Ea/RT

where A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

To calculate k at 75°C, we need to find the frequency factor (A) for the reaction. Unfortunately, the question does not provide the value of A, so we cannot calculate the rate constant at 75°C.

The rate constant ( k ) at 75°C is approximately [tex]\( {2.57 \times 10^{-5} \text{ s}^{-1}} \).[/tex]

To find the rate constant ( k ) at 75°C for a reaction with an activation energy given at 25°C, we'll use the Arrhenius equation. The Arrhenius equation relates the rate constant ( k ) to temperature ( T ) and the activation energy [tex]\( E_a \):[/tex]

[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]

Given data:

Step-by-Step Solution:

1. Convert activation energy to Joules per mole:

[tex]\[ E_a = 33.6 \times 10^3 \text{ J/mol} \][/tex]

2. Calculate ( k ) at 25°C (298.15 K):

First, express the Arrhenius equation in terms of [tex]\( k_{25} \)[/tex] and solve for ( A ):

[tex]\[ k_{25} = A \cdot e^{-\frac{E_a}{RT_{25}}} \][/tex]

[tex]\[ A = k_{25} \cdot e^{\frac{E_a}{RT_{25}}} \][/tex]

Calculate ( A ):

[tex]\[ A = 4.7 \times 10^{-3} \text{ s}^{-1} \cdot e^{\frac{33.6 \times 10^3 \text{ J/mol}}{8.314 \text{ J/(mol·K)} \cdot 298.15 \text{ K}}} \][/tex]

[tex]\[ A \approx 4.7 \times 10^{-3} \text{ s}^{-1} \cdot 6.55 \times 10^5 \][/tex]

[tex]\[ A \approx 3.08 \text{ s}^{-1} \][/tex]

3. Calculate ( k ) at 75°C (348.15 K):

Now use the calculated ( A ) and the new temperature [tex]\( T_{75} = 348.15 \) K:[/tex]

[tex]\[ k_{75} = A \cdot e^{-\frac{E_a}{RT_{75}}} \][/tex]

[tex]\[ k_{75} = 3.08 \text{ s}^{-1} \cdot e^{-\frac{33.6 \times 10^3 \text{ J/mol}}{8.314 \text{ J/(mol·K)} \cdot 348.15 \text{ K}}} \][/tex]

[tex]\[ k_{75} = 3.08 \text{ s}^{-1} \cdot e^{-11.51} \][/tex]

[tex]\[ k_{75} = 3.08 \text{ s}^{-1} \cdot 8.35 \times 10^{-6} \][/tex]

[tex]\[ k_{75} \approx 2.57 \times 10^{-5} \text{ s}^{-1} \][/tex]

Answer:

23 kPa = Partial pressure O₂

Explanation:

In a mixture of gases, the sum of partial pressure of each gas that contains the mixture = Total pressure

Total pressure = Partial pressure N₂ + Partial pressure CO₂ + Partial pressure O₂

95 kPa = 48 kPa + 24 kPa + Partial pressure O₂

95 kPa - 48 kPa - 24 kPa = Partial pressure O₂

23 kPa = Partial pressure O₂

Answer:

The answer to your question is 8.67 grams of Al₂O₃

Explanation:

Data

mass of Al₂O₃ = ?

mass of Al = 4.6 g

mass of O₂ = excess

Balanced Reaction

                   4Al   +    3O₂     ⇒    2Al₂O₃

            Reactants           Elements           Products

                    4                     Al                          4

                    6                      O                         6

Process

1.- Use proportions to calculate the moles of Al

                         27 g of Al ------------------  1 mol

                         4.6 g of Al ------------------  x

                           x = 0.17 mol of Al

2.- use proportions to calculate the moles of Al₂O₃

                   4 moles of Al ------------------  2 moles of Al₂O₃

                   0.17 moles of Al --------------  x

                        x = 0.085 moles of Al₂O₃

3.- Use proportions to calculate the grams of Al₂O₃

molecular mass Al₂O₃ = (27 x 2) + (16 x 3) = 102 g

                     102 g of Al₂O₃ ---------------  1 mol

                       x                     --------------- 0.085 moles

                       x = 8.67 g of Al₂O₃

Answer:

The molarity is 1.33×10^-7M

Explanation:

15ppb = 15g of chlorobenzene/10^9g of solution × 1 mole of chlorobenzene/112.6g × 1g of solution/mL = 1.33×10^-10mol/mL × 1000mL/1L = 1.33×10^-7mol/L

The molarity of a chlorobenzene solution with a concentration of 15 ppb is calculated by first converting the ppb measurement to mass per volume and then using the molar mass to find the number of moles per liter, resulting in a molarity of 1.33 × 10⁻⁷ M.

To calculate the molarity of chlorobenzene in a solution with a concentration of 15 ppb, we first need to convert this concentration into a mass-volume relationship. Since 1 ppb is equivalent to 1 µg/L, 15 ppb is equal to 15 µg/L, which is the same as 0.015 mg/L.

Using the molar mass of chlorobenzene, which is 112.6 g/mol, we can convert this mass into moles using the following equation:

The solution has a concentration of 15 µg/L, which is equal to 0.015 mg/L or 0.000015 g/L. So the amount of moles of chlorobenzene is:

Number of moles = 0.000015 g / 112.6 g/mol = 1.33 × 10-7 moles/L

Therefore, the molarity of the chlorobenzene solution is 1.33 × 10-7 M.

Answer:

The correct answer is a. Chloride

Explanation:

GABAA receptors, is an ionotropic receptor that controls most of the central nervous system inhibitory transmission. GABAA receptors exist as ionotropic ligand-gated ion channel and selectively conduct chloride ions Cl⁻ when activated by GABA through its pore where by the flow of Cl⁻ depends on the internal voltage of the cell and the resting potential. Cl⁻ will flow in a cell if the internal voltage is more than resting potential which is -75 mV. When the internal voltage is less than resting potential, Cl⁻ will transit out of the cell

Final answer:

The ion responsible for the changes when GABA is released onto Cell Y with a resting membrane potential at -55mV and only GABAA receptors present is chloride, which causes an inhibitory postsynaptic potential.

Explanation:

When Cell X fires an action potential and releases GABA onto Cell Y, and only GABAA receptors are present on the postsynaptic membrane which is at rest at -55mV, the ion responsible for the observed changes is chloride. The binding of GABA to GABAA receptors increases the influx of chloride ions into the postsynaptic cell, making the inside of the cell more negative. This increase in negative charge pulls the membrane potential towards the equilibrium potential of chloride which is -65 mV and creates an inhibitory postsynaptic potential (IPSP), thus inhibiting the neuron from firing an action potential.

Answer : The rate law for the overall reaction is, [tex]Rate=k[NO]^2[H_2][/tex]

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : [tex]2NO+H_2\rightarrow N_2+H_2O_2[/tex]    (slow)

Step 2 : [tex]H_2O_2+H_2\rightarrow 2H_2O[/tex]     (fast)

Overall reaction : [tex]2NO+2H_2\rightarrow N_2+2H_2O[/tex]

The rate law expression for overall reaction should be in terms of [tex]NO\text{ and }H_2[/tex].

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

[tex]2NO+H_2\rightarrow N_2+H_2O_2[/tex]

The expression of rate law for this reaction will be,

[tex]Rate=k[NO]^2[H_2][/tex]

Hence, the rate law for the overall reaction is, [tex]Rate=k[NO]^2[H_2][/tex]

Answer:

the initial amount of calcium oxide will decrease as the system moves toward equilibrium

Explanation:

The question should have this variable:

a) 655g CaCO3, 95.0g CaO, P(CO2) = 2.55atm

The reaction formula of dry ice, calcium oxide and calcium carbonate should be:

CaCO3 (s) --> CaO(s) + CO2(g)

With Kp = 1.04 at 900 degrees C

in this tank, only CO2(g) is in gaseous form. The amount of gas will determine the pressure of the tank. When the tank heated, its temperature rise and will also rise the pressure. When pressure higher than Kp, the equilibrium will shift toward side with less gas.

There is only one gas in the reaction, so Kp=  P(CO2)

Since P(CO2) is 2.55 atm and its higher than Kp(1.04), the equilibrium will shift to the left. Since the reaction direction to the left, the amount of calcium carbonate will increase while carbon dioxide and calcium oxide will decrease.

The initial amount of calcium oxide will decrease because  the system moves toward equilibrium

In this tank, only CO2(g) should be in gaseous form. The amount of gas will measure the pressure of the tank. At the time When the tank is heated, its temperature should be increased due to this the pressure is also increased.  

Since P(CO2) is 2.55 atm and it's more than Kp(1.04), the equilibrium should shift to the left. The reaction shifted to the left, the amount of calcium carbonate will increase while on the other hand, carbon dioxide and calcium oxide will decrease.

learn more about volume here: https://brainly.com/question/22083481

The average rate of the reaction between 20 and 30 seconds is 0.012 M/s.

To calculate the average rate, we need to know the change in concentration of the reactant (A) and the time interval over which the change occurred. In this case, the change in concentration of A is 0.012 M (from 0.06 M to 0.048 M) and the time interval is 10 seconds (from 20 seconds to 30 seconds).

The average rate of the reaction is calculated as follows:

Average rate = Change in concentration / Time interval

= (0.012 M) / (10 s)

= 0.012 M/s

Therefore, the average rate of the reaction between 20 and 30 seconds is 0.012 M/s.

To learn more about rate of the reaction, here

https://brainly.com/question/28566775

#SPJ2

The complete question is:

This graph shows the concentration of the reactant Ain the reaction A→B. Determine the average rate of the reaction between 20 and 30 seconds.

Final answer:

To calculate the percent composition of aluminum and magnesium in the sample, we use stoichiometry. Moles of hydrogen gas produced help to establish moles of the metals reacting, but without complete reaction equations for the alloy, or more data, a definitive calculation cannot be provided.

Explanation:

To calculate the percentage by mass of aluminum and magnesium in the sample, we need to perform a few stoichiometric calculations based on the reaction of the alloy with hydrochloric acid and the production of hydrogen gas. The typical reactions for aluminum and magnesium with hydrochloric acid are:

Using the molar mass of hydrogen (1.008 g/mol), we can calculate the moles of hydrogen gas produced using the equation:

moles H₂ = 0.998 g / (2 * 1.008 g/mol) = 0.495 moles

The mole ratio of Al to H₂ in the reaction is 2:3, and for Mg to H₂ is 1:1. We can calculate the mass of aluminum or magnesium that would produce 0.495 moles of hydrogen.

For aluminum:

moles Al = (2/3) * moles H₂ = (2/3) * 0.495 = 0.330 moles

For magnesium:

moles Mg = moles H₂ = 0.495 moles

Now, we calculate the mass:

mass Al = moles Al * atomic mass Al = 0.330 moles * 26.98 g/mol = 8.904 g

mass Mg = moles Mg * atomic mass Mg = 0.495 moles * 24.31 g/mol = 12.033 g

These numbers are hypothetical maximums if the sample was 100% Al or Mg, and do not add up to 9.87 g. Therefore, we need to set up a system of equations considering the total mass of the alloy and the mass of hydrogen produced to find the exact mass of Al and Mg in the sample. This requires more information or a different approach that accounts for the actual reaction stoichiometry with the alloy sample, which isn't provided here.

The alloy consists of: 62.2% aluminium, 37.8% magnesium

To calculate the percentage by mass of each metal (aluminium and magnesium) in the alloy, we need to use the stoichiometry of their reactions with hydrochloric acid and the information given about the mass of the hydrogen gas produced.

Step 1: Write the balanced chemical equations for the reactions

For aluminium:

[tex]\[ 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \][/tex]

For magnesium:

[tex]\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \][/tex]

Step 2: Determine the moles of hydrogen gas produced

The molar mass of hydrogen gas [tex](\(\text{H}_2\))[/tex] is:

[tex]\[ \text{Molar mass of H}_2 = 2 \times 1.008 = 2.016 \, \text{g/mol} \][/tex]

The number of moles of hydrogen gas produced is:

[tex]\[ \text{Moles of H}_2 = \frac{0.998 \, \text{g}}{2.016 \, \text{g/mol}} \]\[ \text{Moles of H}_2 = 0.495 \, \text{mol} \][/tex]

Step 3: Relate moles of hydrogen gas to moles of metals

Let  x be the mass of aluminium and  y be the mass of magnesium in the alloy.

From the reactions:

- Aluminium produces 3 moles of [tex]\(\text{H}_2\)[/tex] per 2 moles of [tex]\(\text{Al}\)[/tex]:

[tex]\[ 2\text{Al} \rightarrow 3\text{H}_2 \]\[ \frac{3}{2} \text{moles of H}_2 \text{ per mole of Al} \]\[ 1 \text{ mole of Al} \rightarrow \frac{3}{2} \text{ moles of H}_2 \][/tex]

- Magnesium produces 1 mole of [tex]\(\text{H}_2\) per mole of \(\text{Mg}\):[/tex]

[tex]\[ \text{Mg} \rightarrow \text{H}_2 \]\[ 1 \text{ mole of Mg} \rightarrow 1 \text{ mole of H}_2 \][/tex]

Thus, the moles of hydrogen gas produced by aluminium and magnesium are:

[tex]\[ \text{Moles of H}_2 \text{ from Al} = \frac{3}{2} \times \frac{x}{26.98} \]\[ \text{Moles of H}_2 \text{ from Mg} = \frac{y}{24.305} \][/tex]

The total moles of hydrogen gas produced:

[tex]\[ \frac{3}{2} \times \frac{x}{26.98} + \frac{y}{24.305} = 0.495 \][/tex]

Step 4: Set up the mass equation for the alloy

The total mass of the alloy is:

[tex]\[ x + y = 9.87 \, \text{g} \][/tex]

Step 5: Solve the system of equations

We have two equations:

[tex]1. \[ \frac{3}{2} \times \frac{x}{26.98} + \frac{y}{24.305} = 0.495 \]\\[/tex]

2.  x + y = 9.87

Let's solve these equations step by step.

First, express  y  in terms of x  from the second equation:

[tex]\[ y = 9.87 - x \][/tex]

Substitute this into the first equation:

[tex]\[ \frac{3}{2} \times \frac{x}{26.98} + \frac{9.87 - x}{24.305} = 0.495 \][/tex]

Simplify and solve for x :

[tex]\[ \frac{3x}{2 \times 26.98} + \frac{9.87 - x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} + \frac{9.87 - x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} + \frac{9.87}{24.305} - \frac{x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} - \frac{x}{24.305} = 0.495 - \frac{9.87}{24.305} \][/tex]

Calculate the constant term:

[tex]\[ 0.495 - \frac{9.87}{24.305} = 0.495 - 0.406 = 0.089 \][/tex]

Combine the \( x \)-terms:

[tex]\[ \frac{3x}{53.96} - \frac{x}{24.305} = 0.089 \]\[ x \left( \frac{3}{53.96} - \frac{1}{24.305} \right) = 0.089 \][/tex]

Calculate the coefficient of \( x \):

[tex]\[ \frac{3}{53.96} = 0.0556 \]\[ \frac{1}{24.305} = 0.0411 \]\[ 0.0556 - 0.0411 = 0.0145 \][/tex]

Now, solve for x :

[tex]\[ x \times 0.0145 = 0.089 \]\[ x = \frac{0.089}{0.0145} \]\[ x = 6.14 \, \text{g} \][/tex]

Now, find y :

[tex]\[ y = 9.87 - x \]\[ y = 9.87 - 6.14 \]\[ y = 3.73 \, \text{g} \][/tex]

Step 6: Calculate the percentage by mass of each metal

[tex]\[ \% \, \text{Al} = \frac{6.14 \, \text{g}}{9.87 \, \text{g}} \times 100 = 62.2\% \]\[ \% \, \text{Mg} = \frac{3.73 \, \text{g}}{9.87 \, \text{g}} \times 100 = 37.8\% \][/tex]

Answer:

The correct answer is "False".

Explanation:

The Elizabethan Era was a period of England's history that corresponds to  the reign of Queen Elizabeth I (1558–1603). The Elizabethan Era is considered a golden age of England's history, and part of the Renaissance Era (1300-1600). The Elizabethan view of life changed drastically from  the characteristic Medieval view of life. During Medieval times life was seen with a religiously perspective only, while during The Elizabethan Era more people start to view life with a scientific perspective.

THE ANSWER IS FALSE!

Answer:the same as the ionization energy trend but opposite the atomic radius trend

Explanation:

Electron affinity refers to the ability of an atom to accept electrons and form a negative ion. This ability increases across the period but decreases down the group. The atomic radius of elements decrease across the period as more nuclear charge is added without a corresponding increase in the number of shells. As size of the nuclear charge increases, the ionization also increases. Down the group, the addition of more shells increases the distance of the outermost electron from the nucleus hence ionization energy decreases and atomic size increases. Electron affinity has the same trend as ionization energy but an opposite trend to atomic radius hence the answer.

The electron affinity trend is B. the same as the ionization energy trend but opposite the atomic radius trend.

Electron affinity simply means the ability of an atom to be able to accept electrons and then form a negative ion. It should be noted that an electron affinity trend increases across the period but decreases down the group.

On the other hand, the atomic radius of elements will decrease across the period when there are more nuclear charges that are added without a corresponding increase in the number of shells. Therefore, as the size of the nuclear charge increases, the ionization also increases.

In conclusion, the electron affinity trend is the same as the ionization energy trend but opposite the atomic radius trend.

Read related link on:

https://brainly.com/question/23558198

Answer:

1.63425 × 10^- 18 Joules.

Explanation:

We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.

We will be using the formula in the equation (1) below;

Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).

Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.

Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;

∆E= - R(H) × [1/nf^2 - 1/ni^2 ].

Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.

Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].

= -2.179 × 10^-18 Joules × (0.25 - 1).

= - 2.179 × 10^-18 × (- 0.75).

= 1.63425 × 10^- 18 Joules.

Answer:

Molarity of acid, Ca = Cb*Vb*A/Va*B

Explanation:

Using H2SO4 as acid, the reaction is as follow:

2NaOH  +  H2SO4 ⇒ Na2SO4  +  2H2O

Volume of acid = Va; Volume of base = Vb, Molar concentration of  acid = Ca; Molar concentration of base = Cb; Molarity of acid = A and Molarity of base = B

Ca*Va/Cb*Vb =A/B

∴ Ca = Cb*Vb*A/Va*B

Answer: C

Explanation:

Pesticide, a substance used for destroying insects or other organisms harmful to cultivated plants or to animals.

Answer:

Answer: C

Explanation:

Pesticide, a substance used for destroying insects or other organisms harmful to cultivated plants or to a

Answer:The stoichiometric coefficient for cobalt (Co) is 1.

Explanation:

To balance the chemical equation for the reaction between aluminum metal (Al) and aqueous cobalt(II) nitrate [Co(NO3)2], you need to ensure that the number of atoms of each element is the same on both sides of the equation. The given reaction can be written as follows:

Al + Co(NO3)2 → Al(NO3)3 + Co

Now, let's balance the equation:

Balance the aluminum (Al) atoms:

There is 1 Al atom on the left and 1 Al atom on the right. Aluminum is already balanced.

Balance the cobalt (Co) atoms:

There is 1 Co atom on the left, but 1 Co atom on the right. Cobalt is already balanced.

Balance the nitrogen (N) atoms:

There are 2 nitrate ions (NO3-) on the left and 3 nitrate ions on the right (since there are three nitrate ions in Al(NO3)3). To balance the nitrogen atoms, we need to put a coefficient of 3 in front of Co(NO3)2 on the left:

Al + 3Co(NO3)2 → Al(NO3)3 + Co

Now, the nitrogen atoms are balanced.

Balance the oxygen (O) atoms:

On the left, there are 3 nitrate ions, which contribute 3 x 3 = 9 oxygen atoms. On the right, there are 3 nitrate ions and 3 oxygen atoms in Al(NO3)3, which contribute a total of 3 x 3 + 3 = 12 oxygen atoms. To balance the oxygen atoms, we need to put a coefficient of 3 in front of Al(NO3)3 on the right:

Al + 3Co(NO3)2 → 3Al(NO3)3 + Co

For more questions related to stoichiometric coefficients

https://brainly.in/question/372962

#SPJ3

Answer:

They tend to have the amide nitrogen protonated to give a positive charge.

Explanation:

A peptide bond joins two consecutive amino acids in the protein. The peptide bond is present between -CO group (also known as carboxyl group) of one amino acid and -NH2 group (also known as amino group) of another amino acid. It is represented as -CONH bond. Therefore, it is an amide linkage. The peptide bond always has planar orientation with trans configuration.  

Trans configuration avoids steric hindrance and hence, add to stability of the peptide bond.  

Nitrogen atom of peptide bond never bear positive charge

Therefore, the incorrect statement is as follows:

They tend to have the amide nitrogen protonated to give a positive charge.

Answer:

2

Explanation:

A careful look at CO2, reveals that CO2 contains:

1 atom of C

2 atoms of O.

Answer:

2

Explanation:

i just took the test

Answer: 0,4278g of F and 0,4191g of Fe

Explanation: it's possible to calculate the mass of each element by multiplying the percentage (decimal) of the element by the mass of the compound.

For Fluorine (F)

0,847g * 0,5051 = 0,4278g of F

For iron (Fe)

0,847 * 0,4949 = 0,4191g of Fe

This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.

At the end, the sum of the elements masses should be the total mass of the compound.

Answer:fluorine=0.5082

Iron=0.3388

Explanation:

Using Empirical formula to show the ratio.

F. Fe

50.51/39 49.49/56

=1.295. 0.88375

=1.295/0.88375 :0.88375/0.88375

=1.465:1

multiply each term by 2 to get a whole number ratio,we have

=(1.465*2) :(1*2)

=2.93:2

=3:2

To get the amount if each contribution of F and Fe,we use ratio,

F=3/(3+2)=3/5*total mass(0.847)

F=0.5082g

Similarly,Fe=2/5*0.847

Fe=0.3388g.

Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is -162.5 kJ/mol

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

[tex]O_3(g)+Cl(g)\rightarrow ClO(g)+O_2[/tex]      [tex]\Delta H^o_{rxn}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]ClO(g)+O_3(g)\rightarrow Cl(g)+2O_2(g)[/tex]    [tex]\Delta H_1=-122.8kJ/mol[/tex]

(2) [tex]2O_3(g)\rightarrow 3O_2(g)[/tex]     [tex]\Delta H_2=-285.3kJ/mol[/tex]

The expression for enthalpy of the reaction follows:

[tex]\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[1\times \Delta H_2][/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(1\times (-(-122.8))+(1\times (-285.3))=-162.5kJ/mol[/tex]

Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is -162.5 kJ/mol

Answer : The molar enthalpy of reaction is, 33.3 KJ/mole

Explanation :

First we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = [tex]60ml+60ml=120ml[/tex]

Now we have to calculate the heat absorbed during the reaction.

[tex]q=m\times c\times (\Delta T)[/tex]

where,

q = heat absorbed = ?

[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

m = mass of water = 120 g

[tex]\Delta T[/tex] = change in temperature = [tex]5.18^oC[/tex]

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

[tex]q=120g\times 4.18J/g^oC\times (5.18)^oC[/tex]

[tex]q=2598.288J=2.60KJ[/tex]

Now we have to calculate the molar enthalpy of reaction.

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy of neutralization = ?

q = heat released = 2.60 KJ

n = number of moles =

[tex]\Delta H=\frac{2.60KJ}{0.078mole}=33.3KJ/mole[/tex]

Therefore, the molar enthalpy of reaction is, 33.3 KJ/mole

The molar enthalpy of reaction of HCl(aq) is calculated to be approximately -33.28 kJ/mol, using calorimetry principles and the given data.

To calculate the molar enthalpy of reaction of HCl(aq), we use the concept of calorimetry and the given temperature change. The provided reaction is a neutralization reaction where HCl reacts with AgNO₃ to form AgCl(s) and HNO₃(aq)

Firstly, we need to calculate the total heat (q) released during the reaction. This can be done by using the formula:

q = mcΔT

where m is the mass of the solution, c is the specific heat capacity of water (assumed to be 4.18 J/g°C as the specific heat capacity is not provided), and ΔT is the change in temperature. The mass (m) of the solution is the sum of the volumes of AgNO₃(aq) and HCl(aq) solutions, assuming a density of 1 g/mL for both solutions:

m = volume of AgNO₃ + volume of HCl = 60 mL + 60 mL

= 120 mL

= 120 g

Substitute the values into the formula to calculate q:

q = (120 g)(4.18 J/g°C)(5.18°C)q

= 2596.416 J

= 2.596 kJ

Next, calculate the number of moles of HCl which is equal to the moles of AgNO₃ since they react in a 1:1 ratio:

Moles of HCl = Volume of HCl × Concentration of HCl

= 0.060 L × 1.30 mol/L

= 0.078 mol

Finally, divide the total heat energy by the number of moles to find the molar enthalpy (ΔH) of the reaction per mole of HCl:

ΔH = q / moles of HClΔH

= 2.596 kJ / 0.078 mol

= -33.28 kJ/mol

Therefore, the molar enthalpy of reaction of HCl(aq) is approximately -33.28 kJ/mol.