The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????,k, is 6.1×10−8 s−1.6.1×10−8 s−1. What is the value of the rate constant at 785.0 K?

The equilibrium concentration of H2O(g) in the given reaction is approximately 12.91 × 10^-6 M.

To determine the equilibrium concentration of H2O(g) in the given reaction, we need to use the equilibrium constant expression and the equilibrium concentrations of the other species. The equilibrium constant expression for the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g) is Kc = [C2H5OH]/([C2H4][H2O]). Given that [C2H4]eq = 0.015 M and [C2H5OH]eq = 1.69 M, we can substitute these values into the expression to solve for the equilibrium concentration of H2O(g).

Kc = [C2H5OH]/([C2H4][H2O])

9.0 × 10^3 = 1.69/((0.015)([H2O]))

[H2O] = 1.69/((0.015)(9.0 × 10^3))

[H2O] ≈ 12.91 × 10^-6 M

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The equilibrium concentration of H2O(g) is approximately 0.0125 M.

The equilibrium concentration of H2O(g) given the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g), the equilibrium constant (Kc), and the equilibrium concentrations of C2H4 and C2H5OH at a particular temperature. We start by writing the expression for the equilibrium constant:

Kc = [C2H5OH]/[C2H4][H2O]

Then, we plug in the known values:

9.0 × 10^3 = 1.69/[0.015][H2O]

Now, we solve for [H2O]:

[H2O] = 1.69/(9.0 × 10^3 × 0.015)

[H2O] ≈ 1.69/(135) ≈ 0.0125 M

Answer:

0.185moles of Al₂O₃

Explanation:

Mass of Al = 10g

Mass of O₂ = 19g

Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃

This is the balanced reaction equation.

Solution

From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.

Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.

Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:

 Number of moles of Al = [tex]\frac{mass}{molar mass}[/tex]

                                        =   [tex]\frac{10}{27}[/tex]

                                        = 0.37mol

From the equation:

         4 moles of Al produced 2 moles of Al₂O₃

    0.37 mole will yield:  [tex]\frac{2 x 0.37}{4}[/tex] = 0.185moles of Al₂O₃

0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.

The equation of the reaction between aluminium and oxygen gas to produce aluminium oxide is given below as follows:

[tex]4Al + 3O₂ → 2Al₂O₃[/tex]

From the equation of reaction 4 moles of aluminium reacts with 3 moles of oxygen to produce 2 moles of aluminium oxide.

The moles of reactants is calculated using:

molar mass of aluminium = 27.0 g

molar mass of oxygen = 32.0 g

moles of aluminium = 10/27 = 0.37 moles

moles of oxygen = 19/32 = 0.59

Aluminium is the limiting reactant

Thus, moles of aluminium oxide produced = 0.37 × 2 = 0.74 moles

Therefore, 0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.

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First we need to calculate the number of moles of FeS[tex]_{2}[/tex]:

number of moles = mass (grams) / molecular mass (g/mol)

number of moles of FeS[tex]_{2}[/tex] = 198.2/120 = 1.65 moles

From the chemical reaction we deduce that:

if            4 moles of FeS[tex]_{2}[/tex] produces 8 moles of SO[tex]_{2}[/tex]

then  1.65 moles of FeS[tex]_{2}[/tex] produces X moles of SO[tex]_{2}[/tex]

X = (1.65×8)/4 = 3.3 moles of SO[tex]_{2}[/tex]

Now we can calculate the mass of SO[tex]_{2}[/tex]:

mass (grams) = number of moles × molecular mass (grams/mole)

mass of SO[tex]_{2}[/tex] = 3.3×64 = 211.2 g

The isotope symbol for the barium (Ba) nucleus in the given reaction is 13756Ba. The isotope symbol for the strontium (Sr) nucleus in the second reaction is 8838Sr.

In the given nuclear fission reaction, the isotope symbol for the barium (Ba) nucleus is 13756Ba. This can be determined by balancing the atomic numbers and mass numbers on both sides of the reaction.

In the second reaction, the isotope symbol for the strontium (Sr) nucleus is 8838Sr. Again, this can be obtained by balancing the atomic numbers and mass numbers.

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Answer:

ΔG° =-990.17 kJ /mol

Explanation:

The equation is:

ΔG° = Δ H° -TΔS°  ....................equation 1

Δ H°rxn = Δ H°f (products)- Δ H°f(reactants)

Δ H°rxn = [[2XΔ H°f(SO₂)]+[2Δ H°f(H₂O)]-[2Δ H°f(H₂S)]+[3Δ H°f(O₂)]

Δ H°rxn = [2X(-296.8)+2(-241.8)]-[2X(-20.6)] = -1036

Δ S°rxn = Δ S°f (products)- Δ S°f(reactants)

Δ S°rxn = [[2XΔ S°f(SO₂)]+[2Δ S°f(H₂O)]-[2Δ S°f(H₂S)]+[3Δ S°f(O₂)]

Δ S°rxn = [2(248.2)+2(188.8)]-[2(205.8)+3(205.2)] = -153.8 J/mol K

      = -0.1538 KJ /mol

Putting values in equation 1  ( T = 298 K)

ΔG° = Δ H° -TΔS° = (-1036)-[(298)(-0.1538)

ΔG° =-990.17 kJ /mol

To calculate the standard change in free energy, ΔG°rxn, use the equation: ΔG°rxn = ΔH°rxn - TΔS°rxn. Calculate ΔH°rxn using the enthalpy of formation values and ΔS°rxn using the entropy values. Substitute the calculated values into the equation to find ΔG°rxn.

To calculate the standard change in free energy, ΔG°rxn, we can use the equation: ΔG°rxn = ΔH°rxn - TΔS°rxn. From the given information, we can calculate the values for ΔH°rxn and ΔS°rxn. First, calculate ΔH°rxn by using the enthalpy of formation values: ΔH°rxn = (2 mol × ΔHf° SO2(g) + 2 mol × ΔHf° H2O(g)) - (2 mol × ΔHf° H2S(g) + 3 mol × ΔHf° O2(g)). Then, calculate ΔS°rxn by using the entropy values: ΔS°rxn = (2 mol × S° SO2(g) + 2 mol × S° H2O(g)) - (2 mol × S° H2S(g) + 3 mol × S° O2(g)). Finally, with the calculated values, substitute them into the equation to find the value for ΔG°rxn.

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The value of Q for the reaction A (g) → 2 B (g) at 298 K when ΔG = -20.5 kJ/mol is approximately 0.00069.

To find the value of the reaction quotient Q when ΔG is known, we use the relationship between Gibbs free energy change, the equilibrium constant K, and Q.

The equation is: ΔG = ΔG° + RT ln(Q)

Where:

First, we need to calculate ΔG°. This can be done using the equilibrium constant (K): ΔG° = -RT ln(K)

Given K = 14.7:

Now we substitute ΔG and ΔG° back into the equation: ΔG = ΔG° + RT ln(Q)

-20.5 kJ/mol = -2.475 kJ/mol + (0.008314 kJ/mol·K) x (298 K) x ln(Q)

Solving for ln(Q):

-20.5 + 2.475 = (0.008314 x 298) x ln(Q)

-18.025 = 2.477 x ln(Q)

ln(Q) = -18.025 / 2.477

ln(Q) ≈ -7.28

Finally, taking the exponent to find Q: Q = [tex]e^(^-^7^.^2^8^)[/tex]

   Q ≈ 0.00069

To find the reaction quotient (Q) when ΔG = -20.5 kJ/mol, we calculate ΔG° using the given value of K and the equation ΔG° = -RT ln K. Then, we use the relationship ΔG = ΔG° + RT ln Q to solve for Q, resulting in Q =0.0037.

To determine the reaction quotient (Q) for the reaction A(g) → 2 B(g) at 298 K when ΔG = -20.5 kJ/mol, we use the relationship between ΔG, Q, and K. The equation is ΔG = ΔG° + RT ln Q. We rearrange to find Q: ln Q = (ΔG - ΔG°) / RT, where ΔG° = -RT ln K

First, calculate ΔG°:

Therefore, Q = 0.0037

Answer: T2= 962.2 K

Explanation:

The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.

So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so

V1=V2  ⇒

[tex]\frac{n1RT1}{P1} = \frac{n2RT2}{P2} \\\\\frac{T1}{P1} = \frac{T2}{P2} \\\\T2= \frac{T1P2}{P1} =\frac{170 K 283KPa}{50 KPa} =962.2 K[/tex]

Answer : The correct option is, kilogram of solvent.

Explanation :

Molality : It is defined as the number of moles of solute present in one kilogram of solvent.

Or, we can say that the number of moles of solute per kilogram of solvent.

The unit of molality is mole per kilogram or mole/Kg.

The formula will be :

[tex]\text{Molality}=\frac{\text{Moles of solute}}{\text{Mass of solvent in Kg}}[/tex]

Hence, the molality is a unit of concentration that measures the moles of solute per kilogram of solvent.

Answer:

adding 750 grams of water at 60° Celsius .

Explanation:

Q = m.c.ΔT,

Where, Q is the amount of heat lost by water (Q = ??? J).

m is the mass of water (m = 750.0 g).

c is the specific heat capacity of the water (c = 4.18 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.

Now, we can calculate the Q that is gained by the different added amounts of water:

ΔT = 70.0°C - 50.0°C = 20.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.

ΔT = 70.0°C - 55.0°C = 15.0°C,

∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.

adding 750 grams of water at 60° Celsius

Answer:

Percentage composition of phosphorus is 43.66%.

Percentage composition of oxygen is 56.66%.

[tex]P_2O_5[/tex]is the empirical formula for this compound.

Explanation:

Moles of phosphorus =[tex]\frac{35.07 g}{31 g/mol}=1.1312 mol[/tex]

Moles of phosphorus oxide =[tex]\frac{71.00 g}{31x+16 y g/mol}=z[/tex]

1.1312 moles of moles of phosphorus gives z moles of phosphorus oxide.

The from 1 mol phosphorus :

[tex]\frac{z}{1.1312 }[/tex] moles of phosphorus oxide...(1)

[tex]2xP+yO_2\rightarrow 2P_xO_y[/tex]

According to reaction, 2x moles of phosphorus gives 2 mol of phosphorus oxide.

The from 1 mol phosphorus :

[tex]\frac{2}{2x}[/tex] moles of phosphorus oxide...(2)

(1)=(2)

[tex]\frac{z}{1.1312 }=\frac{2}{2x}=\frac{71.00 g}{31x+16 y g/mol}[/tex]

[tex]x=\frac{2}{5}y[/tex]

The molecular formula of the phosphorus oxide :[tex]P_xO_y[/tex]:

[tex]P_{\frac{2y}{5}}O_{y}=P_2O_5[/tex]

The empirical formula is simplest ratio of elements in a compound.

[tex]P_2O_5[/tex]is the empirical formula for this compound.

Percentage composition of phosphorus =

[tex]P\%=\frac{2\times 31 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]

[tex]P\%=43.66\%[/tex]

Percentage composition of oxygen=

[tex]O\%=\frac{5\times 16 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]

[tex]O\%=56.33\%[/tex]

The percent composition of the phosphorus oxygen compound is 49.4% phosphorus and 50.6% oxygen. The empirical formula, based on the simplest ratio of moles of each component, is P2O4.

Firstly, to obtain the percent composition of the phosphorus oxide compound, we divide the mass of phosphorus by the total mass of the compound, then multiply the quotient by 100. Thus, using the given masses: (35.07 g P/71.00 g P-Oxide) * 100 = 49.4% Phosphorus.

By subtraction, we can find the percentage of Oxygen: 100% - 49.4 % = 50.6% Oxygen. This gives us the percent composition of the phosphorus oxygen compound.

Next, we need to identify the empirical formula from these percentages. We convert percentages to grams (which doesn't change our values because percentages are based on a total of 100), and then we convert from grams to moles using the atomic masses of phosphorus (P) and oxygen (O): 35.07 g P * (1 mol P/30.97 g) = 1.13 mol P and 35.93 g O * (1 mol O/16.00 g) = 2.24 mol O.

To find the empirical formula, we then divide each mole quantity by the smallest obtained mole value (1.13), to generate the simplest mole ratio: P1O1.98, which we round to P1O2 due to natural variation. Therefore, the empirical formula for this phosphorus oxygen compound is P1O2 or more commonly written as P2O4.

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Answer:

Explanation:

1.  The reaction has at least two reactants and one product

Synthesis

Synthesis is a sort of combination reaction in which a single product is formed from  two or more reactants:

               A + B → C

2.  The reaction has one reactant and at least two products

Decomposition

In decomposition reaction, a compound breaks down into individual elements or compounds.

           A → B + C

3.  Oxygen is one of the reactants and a large amount of energy is released

Combustion

In combustion reaction, oxygen combines with a fuel source to produce energy.

4.  One or more atoms replaces a part of a compound

Replacement

In this type of reaction, one substance replaces another we say it is a single replacement reaction. Some other types involves the actual exchange of partners.

5. The products are always salt and water

Neutralization

Neutralization reaction is a reaction in which acids reacts with bases to produce salts and water.

Final answer:

Chemical reactions are matched with their properties: Synthesis is a reaction with multiple reactants and one product, Decomposition has one reactant and multiple products, Combustion involves oxygen and energy release, Replacement consists of atoms substituting part of a compound, and Neutralization results in salt and water.

Explanation:

When matching the chemical reactions in Column B with their properties listed in Column A, the following connections can be made:

Therefore, the correct matches are:

Answer:

Explanation:

1) Reactants:

The reactants are:

2) Stoichiometric coefficients:

        Cl₂ (g) + 3F₂ (g) →

3) Product:

       Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)

        As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.

Answer : The heat released per gram of the compound reacted with oxygen is, 71.915 kJ

Solution :

The balanced chemical reaction is,

[tex]2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{B_2O_3}\times \Delta H_{B_2O_3})]-[(n_{B_5H_9}\times \Delta H_{B_5H_9})+(n_{O_2}\times \Delta H_{O_2})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(9\times -285.83)+(5\times -1271.94)]-[(2\times 73.2)+(12\times 0)]\\\\\Delta H=-9078.57kJ[/tex]

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

As we know that,

1 mole of [tex]B_5H_9[/tex] has 63.12 grams of mass

So, 2 mole of [tex]B_5H_9[/tex] has [tex]2\times 63.12=126.24[/tex] grams of mass

As, 126.24 g of [tex]B_5H_9[/tex] release heat = 9078.57 kJ

So, 1 g of [tex]B_5H_9[/tex] release heat = [tex]\frac{9078.57}{126.24}=71.915kJ[/tex]

Therefore, the heat released per gram of the compound reacted with oxygen is, 71.915 kJ

The heat released per gram of the compound reacted with oxygen is -71.92 KJ/mol per gram of B5H9 reacted.

The equation goes as follows;

2B5H9(l) + 12O2(g) → 5B2O3(s) + 9H2O(l)

We have the following information;

ΔH°f  B5H9(l) =  73.2 kJ/mol

ΔH°f  B2O3(s) = −1271.94 kJ/mol

ΔH°f  H2O(l) = −285.83 kJ/mol

Note that;

ΔHrxn = ∑ΔH°f (products) - ΔH°f (reactants)

ΔHrxn =  ∑(5 × ( −1271.94 kJ/mol)) + (9 × ( −285.83 kJ/mol)) - ∑(2 × (73.2 kJ/mol) + (12 × 0)

ΔHrxn = -9078.57  kJ/mol

Since 1 mole of B5H9 = 63.12 g/mol

Two moles of B5H9 reacted so 2 moles × 63.12 g/mol = 126.24 g

Heat released per gram of B5H9 reacted = -9078.57  kJ/mol/126.24 g

= -71.92 KJ/mol per gram of B5H9 reacted.

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Answer:

2

Explanation:

In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:

   For the mass numbers(superscript):

   235 + 1 = 1 + 139 + 95

       236 = 235

This is not balanced

  For the atomic number:

  92 + 0 = 0 + 53 + 39

      92 = 92

This is balanced.

We simply inspect to see how to balance the mass number.

By putting a coefficient of 2 behind the neutron atom, the equation becomes balanced.

First, in this case, we define the K constant as the solubility of the solute in water divided by the solubility of the solute in ether.

K = (X grams of solute / 75 mL of ethyl ether) / (5 g of solute / 100 mL of water)

K = (X / 75) / (5 / 100)

1 = (X / 75) / (5 / 100)

5 / 100 = X / 75

0,05 = X /75

X = 0,05 × 75 = 3.75 g of solute that will be extracted by the ether

Answer:

The correct answer is 2.44 grams.

Explanation:

The partition coefficient, K = concentration of solute in ether / concentration of solute in water

As partition coefficient is 1, therefore, the concentration of solute in both the solvents would be similar.  

Thus, when 25 milliliters of ether is added to 100 ml of water Wether/25 = Wwater/100

However, Wwater + Wether = 5.00 g

Wwater = 5.00 g - Wether

So, Wether/25 = 5.00 - Wether/100

Wether = 1.00 g

Thus, Wwater = 5.00 - 1.00 = 4.00 grams

So, for the first time, the solute extracted by ether is 1.0 gram

Now add 25 milliliters of ether to 4.00 grams of solute of 100 ml water,

Wwater + Wether = 4.00 g

Wwater = 4.00 g - Wwater

So, Wether/25 = 4.00 - Wether/100

Wether = 0.8

Thus, Wwater = 4.0 - 0.8 = 3.2 grams

So, for the second time the amount of solute extracted by ether is 0.8 gram.  

Now, add 25 ml of ether to 3.2 grams of solute of 100 ml water

Wwater + Wether = 3.20 g

Wwater = 3.20 - Wether

So, Wether/25 = 3.20 - Wether/100

Thus, Wwater = 3.20 - 0.64 = 2.56 grams

So, for the third time, the amount of solute extracted by ether is 0.64 grams.  

Therefore, total weight of solute extracted by ether is 1.00 + 0.80 + 0.64 = 2.44 grams.  

Answer: 160 ml

Explanation:

The expression used will be :

[tex]C_1V_1+C_2V_2=C_3V_3[/tex]

where,

[tex]C_1[/tex] = concentration of Ist alcohol solution= 10%

[tex]C_2[/tex] = concentration of 2nd alcohol solution= 55%

[tex]V_1[/tex] = volume of Ist alcohol solution = 200 ml

[tex]V_2[/tex] = volume of  2nd alcohol solution= v ml

[tex]C_3[/tex] = concentration of resulting alcohol solution= 30%

[tex]V_2[/tex] = volume of resulting alcohol solution= (v+200) ml

Now put all the given values in the above law, we get the volume of added.

[tex](10\times 200)+(55\times v)=(30\times (v+200)ml)[/tex]

By solving the terms, we get :

[tex]v=160ml[/tex]

Therefore, the volume of 55% mixture  needed to be added to obtain the desired solution is 160 ml.

Answer: A) 4C2+S8 -> 4CS2

Explanation:

The chemical equation

4C2+S8 -> 4CS2 is flawed because the equation is not balanced i.e the carbon atom at the reactant is not equal to that of the product and for a chemical reaction to be a balanced equation, the number of atoms of elements at the reactants must be equal to that of the product.

As you can see, there are 8 atoms of carbon(C) at the reactant while we have only 4 carbon atoms at the product therefore making the equation unbalanced.

Answer:

Person one has conduction, person 2 has radiation, and person 3 has convention.

Explanation:

Person 1 is touching the mug to get warm, which is a transfer of heat.

Person 2 is exposed to the fire, which is radiation.

Person 3 is exposed to the warm air of convection.

Explanation:

A process that involves transfer of heat from a hot substance to a cold substance by coming in contact with each other is known as conduction.

For example, a person has his hands cupped around a ceramic mug of hot chocolate then heat is transferring from ceramic mug to the hands.

Whereas a process in which heat energy is transferred in a wave-like motion through the space is known as radiation.

For example, a person is toasting a marshmallow above the fire is getting heat energy in the form of radiation.

Also, third person who is roasting a hot dog above the glowing coals is getting heat energy in the form of radiation.

Hence, we can conclude that primary source of heat transfer for person 1 is conduction, and for both person 2 and 3 is radiation.

Answer: A. The reaction takes place in one step.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

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

[tex]Rate=k[A]^a[B]^b[/tex]

k= rate constant

a= order with respect to A

b = order with respect to B

Answer:

anwer is a

Explanation:

Answer:

The answer is False.

Explanation:

When mixed with an acidic liquid etc. a violent reaction happens creating Hydrogen Gas.

The statement that sodium borohydride is stable in acidic aqueous solutions is false.

Sodium borohydride is a very useful reducing agent in chemistry. It is able to carry out many important organic transformations where reduction reactions are involved.

However, sodium borohydride is unstable in acid solutions. It decomposes in acidic and even neutral solutions to release hydrogen gas. Therefore, the statement that sodium borohydride is stable in acidic aqueous solutions is false.

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Answer:

The activation energy of the reaction is 1.152 kJ/mol.

Explanation:

Activation energy is the minimum amount which is absorbed by the reactant molecules to undergo chemical reaction.

Initial temperature of reaction = [tex]T_1=100 K[/tex]

Final temperature of reaction = [tex]T_2=200 K[/tex]

Initial rate of the reaction at 100 k = [tex]K_1=k[/tex]

Final rate of the reaction at 200 k = [tex]K_2=2k[/tex]

Activation energy is calculated from the formula:

[tex]\log\frac{K_2}{K_1}=\frac{E_a}{2.303\times R}(\frac{T_2-T_1}{T_1T_2})[/tex]

R = Universal gas constant = 8.314 J/ K mol

[tex]\log\frac{2k}{k}=\frac{E_a}{2.303\times 8.314 J/K mol}(\frac{200 K-100K}{200 K\times 100K})[/tex]

[tex]E_a=1,152.772 J/mol=1.152 kJ/mol[/tex]

The activation energy for this chemical reaction can be calculated using the Arrhenius equation and the provided information about the doubling of the rate constant as the temperature increases from 100 K to 200 K. An alternative two-point form of the Arrhenius equation can also be used for this calculation.

The activation energy of a reaction can be calculated by utilizing the Arrhenius equation, which relates the effect of temperature on the rate constant, k of a reaction: k = Ae−Ea/RT. In this equation, R is the ideal gas constant (8.314 J/mol/K), T is the temperature in Kelvin, Ea is the activation energy in joules per mole, A is a constant known as the frequency factor, and e is a mathematical constant.

Given the data that when the temperature increases from 100 K to 200 K the rate constant (k) doubles, we can use an alternative approach to calculate the activation energy. This involves rearranging the Arrhenius equation in a two-point form and inserting the provided temperatures and rate constants. Substitution and subsequent calculation will yield the activation energy value in kJ/mol.

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Answer:

The maximum wavelength of light will be 773.46 nm.

Explanation:

Energy required to break a fluorine-fluorine single bond = 155 kJ/mol

For 1 mole of fluorine gas  = 155 kJ = 155000 J

For [tex]6.022\times 10^{23}[/tex] molecules = 155000 J

Then energy to break 1 molecule = [tex]\frac{155000 J}{6.022\times 10^{23}}=2.57\times 10^{-19} J[/tex]

Energy of the photon E =[tex]2.57\times 10^{-19} J[/tex]

Wavelength of the light =[tex]\lambda [/tex]

[tex]E=\frac{hc}{\lambda }[/tex]

[tex]2.57\times 10^{-19} J=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{\lambda }[/tex]

[tex]\lambda =7.7346\times 10^{-7} m=773.46 nm[/tex]

The maximum wavelength of light will be 773.46 nm.

Answer: The unknown gas is B) [tex]H_2[/tex] .

Explanation: The problem is based on Graham's law of effusion rates of gases.

Carbon dioxide takes 4.69 times as long to escape as the unknown gas. It means the unknown gas is lighter than carbon dioxide since the lighter gas takes less time to escape.

From Graham's law, "Rate of effusion of a gas is inversely proportional to the molar mass of the gas."

Molar mass of carbon dioxide is 44.0 gram per mol. Let's say the molar mass of the unknown gas is M.

[tex]\frac{rate_A}{rate_C_O_2}=\sqrt{\frac{44.0}{M}}[/tex]

[tex]4.69=\sqrt{\frac{44.0}{M}}[/tex]

Do the square to both sides:

[tex]21.9961={44.0}{M}}[/tex]

[tex]M=\frac{44.0}{21.9961}[/tex]

M = 2.00

2.00 gram per mol is the molar mass of hydrogen gas. So, the correct option is B) [tex]H_2[/tex] .

Final answer:

Using Graham's Law of Effusion, the unknown gas effusing 4.69 times faster than carbon dioxide is identified as hydrogen (H2), which has a much lower molar mass, making answer B) H2 the correct choice.

Explanation:

The question involves the application of Graham's Law of Effusion, which relates the rates of effusion of two gases to their molar masses. Given that carbon dioxide takes 4.69 times as long to escape as the unknown gas, we can use the inverse relationship between the rate of effusion and the square root of the molar mass (rate ≈ 1/sqrt(molar mass)) to identify the unknown gas.

The molar mass of carbon dioxide (CO₂) is approximately 44 g/mol.

By setting up a ratio using Graham's Law, (rate of unknown gas/rate of CO₂) = sqrt(molar mass of CO₂/molar mass of unknown gas), and substituting the given rates, we can solve for the molar mass of the unknown gas:

(1/4.69)² = (44 g/mol) / (molar mass of unknown gas)

(1/22.0361) = (44 g/mol) / (molar mass of unknown gas)

Molar mass of unknown gas = 44 g/mol * 22.0361 = 969.8684 g/mol

True, the kilogram is used as a unit for mass in the the international system of units (SI) for scientific measurements.

Answer: True

Explanation:

Mass is defined as the amount of matter contained in the body.

Its units are kg, gram, milligram which are inter convertible.

S.I or M.K.S system has seven fundamental units which are used to find derived units

1) Mass - Kilogram

2) Length - meter

3) Time - Seconds

4) Electric Current - Ampere

5) Amount of substance - Moles

6) Intensity of light - Candela

7) Temperature - Kelvin

Thus the kilogram is the SI unit for mass.

Answer:

Explanation:

A reversible reaction is indicated by using a double arrow ⇄

The upper arrow (→), from left to right,  indicates the direct or forward reaction, which goes from left to right.

In the direct reaction, the reactants are the substances shown on the left side of the equation, and the products are the substances shown on the right side.

The lower arrow (←), from right to left, indicates the reverse reaction, which goes from right to left.

In the reverse reaction, the reactants are the substances shown of the right side and the products are the substances shown of the left side.

Summarizing:

In the moment that the rates of both forward and reverse reactions are equal it is said that the equilibrium has been reached.

Answer:

just did edge test:

c. NH4CI(s)<—–>NH3( g)+HCI(g)

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

[tex]q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ[/tex]

Moles of propane =[tex]\frac{1.860 g}{44 g/mol}=0.0422 mol[/tex]

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

[tex]\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol[/tex]

The simple answer is because of the law of conservation of mass. This law states that matter can not be created or destroyed only transfered or rearranged. You can't have 5 oxygens reacting and only 3 oxygens as the product. What happened to the other two oxygens? They couldn't have magically disappeared. How much you put into an equation is what you will get out of the equation. This is one if the guarantees of life.

Hope this helped!

~Just a girl in love with Shawn Mendes

Answer:

ExplaChemical equations must be balanced to ensure that the number of atoms for each element is equal. Any imbalance would be a violation of the law of conservation of massnation:

Answer:

acetyl CoA

Explanation:

The starting molecule for the krebs cycle is acetyl CoA.

Answer:2500-3300[tex]cm^{-1}[/tex]

Explanation:We can distinguish between 2-butanone and butanoic acid by analyzing the spectra .                                          

The structure of the two compounds are slightly different as butanoic acid has an OH group whereas for 2-butanone there is no OH group present.

Please refer the attachments for structure of compounds.

Both the compounds will show carbonyl stretching frequency at around 1700-1750cm^{-1}[/tex] due to the presence of carbonyl group in both the compounds.

A broad intense peak at around 3000[tex]cm^{-1}[/tex] will be observed only for butanoic acid due to the presence of OH group and 2-butanone would not show any broad intense peak at around 3000[tex]cm^{-1}[/tex].

So by analyzing the IR spectra and identifying the intense broad peak of OH we can easily distinguish between butanoic acid and 2-butanone.

Due to hydrogen bonding in between two carboxylic acid molecules ,peak broadening for OH group in butanoic acid takes place.

Generally the ketones show IR stretching in around 1700-1730cm^{-1}[/tex] due to the presence of carbonyl group.

Generally carboxylic acids show IR stretching  for carbonyl group at around 1700-1760cm^{-1}[/tex] and a broad intense peak for OH at around 2500-3000cm^{-1}[/tex].

The C=O stretching region in the IR spectrum can be used to distinguish between butanoic acid and 2-butanone.

In the IR spectrum, the region that could be used to distinguish between butanoic acid and 2-butanone is the C=O stretching region. The C=O stretching vibration band of saturated aliphatic ketones, such as 2-butanone, appears at around 1715 cm-1. However, butanoic acid contains a carboxyl group (COOH), which exhibits a broad and intense peak in the 1680-1750 cm-1 range.