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2 years ago

PLEASE EXPLAIN I KNOW THE ANSWER I NEED AN EXPLAINATION

Chemistry

1 answer:

BaLLatris [955]2 years ago

6 0

From the graph shown, Y2 = N2, X2 = O2, Z2 = H2.

When atoms of elements are at a large distance from each other, the potential energy of the system is high. However, as the atoms approach each other, the potential energy of the system decreases steadily. The closer the atoms come to each other, the lower the potential energy. This implies that potential energy is proportional to bond length.

Since N2 has a shorter bond length and higher energy, Y2 must be N2. The bond energy of hydrogen is small hence H2 must be X2. The bond energy of a double bond is intermediate between that of a single and triple bond hence Z2 must be O2.

Learn more: brainly.com/question/24857760

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3 years ago

A ray in the emission spectrum has a wavelength of 3.10 x 1014 meters. Given that the speed of light is 2.998 x 108 m/s, what is

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<span>0.967 x 10^-6 HZ

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Give the number of lone pairs around the central atom and the molecular geometry of XeF4.A)0 lone pairs, tetrahedralB)1 lone pai

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Using the VSEPR ( Valence Shell Electron Pair Repulsion)Theory

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The hybridrization of the compound is sp3d2 because the number of electrons around the central atom is 6.

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3 years ago

A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

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Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

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2 years ago