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Vika [28.1K]
4 years ago
10

How many reactant molecules and product gas molecules are in this equation?

Chemistry
1 answer:
Mice21 [21]4 years ago
6 0

Answer:

N₂  = 6.022 × 10²³ molecules

H₂ = 18.066 × 10²³ molecules

NH₃ = 12.044 × 10²³ molecules

Explanation:

Chemical equation;

N₂ + 3H₂     →  2NH₃

It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

Number of molecules of nitrogen gas:

1 mol = 6.022 × 10²³ molecules

Number of molecules of hydrogen:

3 mol × 6.022 × 10²³ molecules/ 1 mol

18.066 × 10²³ molecules

Number of molecules of ammonia:

2 mol × 6.022 × 10²³ molecules/ 1 mol

12.044 × 10²³ molecules

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The first excited electronic energy level of the helium atom is 3.13 x 10-18 J above the ground level. Estimate the temperature
salantis [7]

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸  J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} =   e ^{^{-\dfrac{(\Delta E)}{KT}}}

From the above equation:

Δ E = energy difference =  3.13 × 10⁻¹⁸  J

k = Boltzmann constant = 1.38 × 10⁻²³  J/K

\dfrac{N_2}{N_1} = 0.5

Thus:

0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

-3.00 = -226811.5942 \times \dfrac{1}{T}

T =  \dfrac{-226811.5942}{-3.00 }

T = 75603.86473 K

4 0
3 years ago
Convert the following to number of moles: 4.03 x 10^21 molecules present in AgNO3
castortr0y [4]
Moles = 4.03 x 10^21
             ------------------------
                6.02 x 10^23
 
           =      6.69 x 10^-3 moles.
Hope this helps!


6 0
3 years ago
As the temperature of a liquid increases, the solubility of a gas in the liquid
RoseWind [281]
The correct answer is B. Solubility describes the amount of solute that can be dissolved in a solvent. This value is not constant is affected by many factors. One factor is the temperature. An increase in temperature, a corresponding change in solubility also can be observed. The increase leads to a decrease in the solubility and the opposite. A decrease is observed since gas molecules are now has enough energy to escape the liquid phase and go to the gas phase.
4 0
3 years ago
MARKING BRAINLIEST! please help asap, i need a and b, thank you
defon

The time required to reduce the concentration from 0.00757 M to 0.00180 M is equal to 1.52 × 10⁻⁴ s. The half-life period of the reaction is 9.98× 10⁻⁵s.

<h3>What is the rate of reaction?</h3>

The rate of reaction is described as the speed at which reactants are converted into products. A catalyst increases the rate of the reaction without going under any change in the chemical reaction.

Given the initial concentration of the reactant, C₀= 0.00757 M

The concentration of reactant after time t is C₁= 0.00180 M

The rate constant of the reaction, k = 37.9 M⁻¹s⁻¹

For the first-order reaction: C_t =C_0-kt

0.00180 = 0.00757 - (37.9) t

t =  1.52 × 10⁻⁴ s

The half-life period of the reaction: t_{\frac{1}{2} } =\frac{C_0}{2k}

t_{\frac{1}{2} } =\frac{0.00757}{2\times 37.9}

Half-life of the reaction = 9.98 × 10⁻⁵s

Learn more about the rate of reaction, here:

brainly.com/question/13571877

#SPJ1

5 0
1 year ago
Which statement best describes the energy of activation?
Furkat [3]

activation energy start off chemical reaction

6 0
4 years ago
Read 2 more answers
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