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3 years ago

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What type of bond is H2Se

1 answer:

Vinvika [58]3 years ago

6 0

Answer:The lone pair of electrons takes up more space than a regular bonding pair since it it is not confined to be between two atoms, so it adds coulombic repulsion to the bonding pairs and compresses the angle. Therefore, the bond angle is less than the standard 109.5∘ . It is actually 97.7∘

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1.5 atm is the same pressure as... (1 atm<br> 760 mmHg)

Darina [25.2K]

<h3>Answer:</h3>

1100 mmHg

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Pressure

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.5 atm

[Solve] mmHg

<u>Step 2: Identify Conversions</u>

1 atm = 760 mmHg

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.5 \ atm(\frac{760 \ mmHg}{1 \ atm})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1140 \ mmHg$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

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3 years ago

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The frequency of the radiation is equal to $7.17 \times 10^{19}$ Hertz.

<u>Given the following data:</u>

Photon energy = $4.75 \times 10^{-14}$ Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

$E = hf$

<u>Where:</u>

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f is photon frequency.

Substituting the given parameters into the formula, we have;

$4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}$

Frequency, F = $7.17 \times 10^{19}$ Hertz

Read more: brainly.com/question/16901506

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2 years ago