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3 years ago

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Consider the energy diagram below. xn. mc012-1.jpg Which line indicates a higher reaction rate? A because it has a lower activat

ion energy. B because it has a lower activation energy. A because its mc012-2.jpgGrxn is much lower. B because its mc012-3.jpgGrxn is much lower.

2 answers:

matrenka [14]3 years ago

8 0

Answer:

b

Explanation:

34kurt3 years ago

5 0

Answer is: B because it has a lower activation energy.

For all chemical reaction some energy is required and that energy is called activation energy (energy that needs to be absorbed for a chemical reaction to start), activation energy for reaction B is lower that for reaction A.

Catalysis is the increase in the rate of a chemical reaction due to the participation of an additional substance called a catalyst.

Chemical reactions occur faster with a catalyst because they require less activation energy.

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Determine the a) energy (in eV) and b) wavelength (in cm) corresponding to blue light of frequency 670 THz

Lesechka [4]

Answer :

(a) The energy of blue light (in eV) is 2.77 eV

(b) The wavelength of blue light is $4\times 10^{-5}cm$

Explanation:

The relation between the energy and frequency is:

$Energy=h\times Frequency$

where,

h = Plank's constant = $6.626\times 10^{-34}J.s$

Given :

Frequency = $670THz=670\times 10^{12}s^{-1}$

Conversion used :

$1THz=10^{12}Hz\\1Hz=1s^{-1}\\1THz=10^{12}s^{-1}$

So,

$Energy=(6.626\times 10^{-34}J.s)\times (670\times 10^{12}s^{-1})$

$Energy=4.44\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(4.44\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=2.77eV$

The energy of blue light (in eV) is 2.77 eV

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where,

c = the speed of light = $3\times 10^8m/s$

Frequency = $670\times 10^{12}s^{-1}$

So, Wavelength is:

$670\times 10^{12}s^{-1}=\frac{3\times 10^8m/s}{Wavelength}$

$Wavelength=\frac{3\times 10^8m/s}{670\times 10^{12}s^{-1}}=4\times 10^{-7}m=4\times 10^{-5}cm$

Conversion used : $1m=100cm$

The wavelength of blue light is $4\times 10^{-5}cm$

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3 years ago

Energy is added in the form of heat to a pot of water on the stove. This causes the water to

Monica [59]

The energy would increase

b) increase kinetic energy

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3 years ago

How many liters of nitrogen gas are needed to make 25 mol of nitrogen trifluoride

ikadub [295]

The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;

N₂ + 3F₂ → 2NF₃

The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol

Hence N₂ moles needed = 12.5 mol

At STP (273 K and 1 atm) 1 mol of gas = 22.4 L

Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L

6 0

3 years ago

At STP, fluorine is a gas and iodine is a solid. This observation can be explained by the fact that fluorine has

motikmotik

<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>

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3 years ago

Read 2 more answers

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

<u>The reaction </u>

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) <u>Limiting reagent</u>: $NO$

3) <u>Ammount of excess reagent</u>:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago