Name the following ketone:

Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2 H 2 O 2 ( aq ) ⟶ 2

Ede4ka [16]

Answer:

The value of temperature for unanalysed reaction is 456.1 K

Explanation:

Step 1: Data given

The activation energy of uncatalyzed reaction is 75.0 kJ/mol.

The activation energy of catalyzed reaction is 49.0 kJ/mol.

The temperature for catalyzed reaction is 25 °C

Step 2: The balanced equation

2 H2O2(aq) ⟶ 2H2O(l) + O2(g)

Step 3: According to Arrhenius equation:

k = Ae ^ (-Ea/RT)

⇒ with k = the rate constant

⇒ with Ea = the activation energy = 49 kJ/mol

⇒ with R = the universal gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 298 K

For uncatalyzed reaction, the rate constant is calculated as:

k(uncatalyzed) = Ae ^ (-75000/8.314 J/mol*K * T)

k(uncatalyzed) = Ae ^ (-9020.93/T)

For catalyzed reaction, the rate constant is calculated as:

k(catalyzed) = Ae ^ (-Ea(catalyzed)/RT))

⇒ with Ea(catalyzed) = 49000 J/mol

⇒ with T = 298 K

k(catalyzed) = Ae ^ (-49000J/(8.314J/mol *K * 298K))

k(catalyzed) = Ae ^(-19.78)

k(catalyzed) = k(uncatalyzed)

Ae^(-19.78) = = Ae ^ (-9020.93/T)

-19.78 = -9020.93/T

T = -9020.93/-19.78

T = 456.1 K

The value of temperature for unanalysed reaction is 456.1 K

8 0

3 years ago

Read 2 more answers

Explain how to use the periodic table to deduce the number of protons, neutrons and electrons of an atom of a specific element

Ksju [112]

Answer:

See explanation

Explanation:

The periodic table shows the atomic number and mass number of each element.

We know that the atomic number shows;

The number of protons in the nucleus of the atom
The number of electrons in the neutral atom of the element.

So we obtain the number of protons and electrons by looking at the atomic number shown in the periodic table.

We also know that;

Mass number = Number of protons + number of neutrons

Since number of protons = atomic number of the atom

Number of neutrons = Mass number - atomic number

Hence we obtain the number of protons by subtracting the atomic number from the mass number given in the periodic table.

5 0

3 years ago

In an experiment, a compound was determined to contain 68.94% oxygen and 31.06% of an unknown element by weight. The molecular w

monitta

Answer is: the compound is B₂O₃.

ω(O) = 68.94% ÷ 100%.

ω(O) = 0.6894; percentage of oxygen in the compound.

ω(X) = 31.06% ÷ 100%.

ω(X) = 0.3106; percentage of unknown element in the compound.

If we take 69.7 grams of the compound:

M(compound) = 69.7 g/mol.

n(compound) = 69.7 g ÷ 69.7 g/mol.

n(compound) = 1 mol.

n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.

n(O) = 3 mol.

M(compound) = n(O) · M(O) + n(X) · M(X).

n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.

n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).

8 0

3 years ago

1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.

lora16 [44]

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

5 0

3 years ago

2 years ago a 4.00 l flexible container holds a sample of hydrogen gas at 153 kpa. if the pressure increases to 203 kpa and the

garri49 [273]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

V2 = 153 x 4 / 203

V2 = 3 L

7 0

3 years ago