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snow_tiger [21]
3 years ago
6

It is the gametes formation of a sex cells or reproductive cells.​

Chemistry
2 answers:
mamaluj [8]3 years ago
4 0

Answer:

sana makatulong po sainyo

Effectus [21]3 years ago
3 0

Answer:

Meiosis

Explanation:

.bnsksksksjabsn...

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What do you understand by valency electron and valency shell?​
jasenka [17]

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<h2>total no. of electron present in Valency shell is called valency electron </h2><h2>___________________</h2>

<h2>valency shell is that in which last electron is present</h2>

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Describe use of H₂S an analytical reagent?<br><br>​
Gemiola [76]

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The main use for hydrogen sulfide is in the production of sulfuric acid and elemental sulfur. ... H2S is used to prepare the inorganic sulfides you need to make those products. As a reagent and intermediate, hydrogen sulfide is beneficial because it can prepare other types of reduced sulfur compounds.

7 0
3 years ago
Ion-dipole interactions can occur between any ion and any molecule with a dipole. Identify all of the following pairs of species
Zigmanuir [339]

An ion-dipole interaction is the result of an electrostatic interaction between a charged ion and a molecule that has a dipole. It is an attractive force that is commonly found in solutions, especially ionic compounds dissolved in polar liquids. A cation can attract the partially negative end of a neutral polar molecule, while an anion attracts the positive end of a polar molecule. Ion-dipole attractions become stronger as the charge on the ion increases or as the magnitude of the dipole of the polar molecule increases.

This force of attraction is between an ion and a charge , it is weaker force than covalent bond and ionic bond . EX - The ion dipole interaction takes place between water and sodium ion , in it there is a small charge on oxygen molecule in water which is attracted by sodium charge .

Most commonly found in solutions. Especially important for solutions of ionic compounds in polar liquids.

A positive ion (cation) attracts the partially negative end of a neutral polar molecule.

to learn more about dipole interactions:-

https://brainly.in/question/1157107

4 0
2 years ago
Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict wheth
Elena-2011 [213]

Answer:

\Delta S_{surr} = + 0.32113\: kJ/K

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at <em>constant pressure</em> is given by,

\Delta S_{surr} = \frac{-\Delta H}{T}

\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K}  \right ) = + 0.32113\: kJ/K > 0

<u><em>In the given reaction:</em></u>

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

\therefore \Delta S_{system} <  0

As we know, <em>for a spontaneous process, that the total entropy should be positive.</em>

\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0  

<u>Therefore, at the given temperature,</u>

  • if \Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0 then the given reaction is spontaneous
  • if \Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0 then the given reaction is non-spontaneous
6 0
3 years ago
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