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SVETLANKA909090 [29]
3 years ago
6

Which type of bond is present in hydrogen sulfide (H2S)? The table of electronegativities is given

Chemistry
2 answers:
LuckyWell [14K]3 years ago
7 0

Bonds formed between atoms can be classified as ionic and covalent

Ionic bonds are formed between atoms that have a high difference in the electronegativity values.

In contrast, bonds formed between atoms that have a difference in electronegativity lower than the ionic counterparts are polar covalent bonds.  If the atoms have very similar electronegativities, they form non-polar covalent bonds.

In H2S, the S atom is bonded to 2 H atoms. The electronegativity of H = 2.2 and S= 2.56. Since the difference is not high the bond formed will be covalent (polar covalent).

Alexxandr [17]3 years ago
4 0

Answer:

nonpolar covalent

Explanation:

is answer on plato

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How many moles of neon occupy a volume of 14.3 l at stp? how many moles of neon occupy a volume of 14.3 l at stp? 1.57 moles 0.6
andriy [413]
The  number   of  neon  moles  that  occupy  a volume of 14.3 l  at STP is calculated as follows

At STP 1  mole = 22.4 liters

what about  14.3 liters

by cross  multiplication
= (1 mole x 14.3 l)/22.4 l =0.638  moles  of  neon


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Answer:

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Explanation:

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3 years ago
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iVinArrow [24]

Answer:

The number, such as 98.7 FM, of a radio station represents:

  • <u>the frequency in which is transmitted the radio signal</u>.

Explanation:

<em>The radio FM is the modulated frequency, which means that all the information is sent by just a signal, with different frequencies which difference them</em>, <u>the radio FM use the frequencies from 88 MHz until 108 Mhz (MHz is a measuring unit for the frequency), with a minimal space among them of 0.2 MHz</u>, this last means that you could find a signal in 88.0 MHz, and the next should be 88.2 MHz, next 88.4 MHz and so (at least, regularly the space between two frequencies is more than 0.2 MHz).

5 0
3 years ago
Identify the reactants in the following equation.
Ratling [72]

Answer:

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Explanation:

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3 0
2 years ago
The equilibrium constant for the reaction 2NO2(g) N2O4(g) is Keq = . If a sample at equilibrium was found to contain 0.058 M NO2
Sladkaya [172]

Answer:

\boxed{3.6}

Explanation:

                  2NO₂ ⇌ N₂O₄

E/mol·L⁻¹:   0.058     0.012

K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\

\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}

7 0
3 years ago
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