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Alexandra [31]
3 years ago
7

Empirical formula for S3O9

Chemistry
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

* The empirical formula of a compound shows the ratio of elements present in a compound

* The molecular formula of a compound shows how many atoms of each element are present in a molecule of a compound.

Example: the compound butene has a molecular formula of C4H8. The empirical formula

of butene is CH2 because there is a 1:2 ratio of carbon atoms to hydrogen atoms.

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Assign an oxidation number to each element in the reaction. CO(g) + 2H2(g) → CH3OH(g) In CO, the oxidation number of C is , and
Elan Coil [88]

Answer:

In CO, the oxidation number of C is (+2), and that of O is (-2).

In H_2H2 , the oxidation number of H is (0).  

In CH_3OHCH3OH , the oxidation number of C is (-2), that of O is (-2) and that of H is (+1).

Explanation :

Oxidation number : It is also called oxidation state. It is a number assigned to an element in a chemical compound that represent the number of electrons gained or lost by an element in a compound.

8 0
3 years ago
Read 2 more answers
A gas-forming reaction produces 1.15 m 3 of gas against a constant pressure of 102.0 kPa. Calculate the work done by the gas in
Aleonysh [2.5K]

Answer:

The work done by the gas is 117,300 Joules.

Explanation:

Pressure applied on the gas , P= 102.0 kPa = 102000 Pa ( 1kPa = 1000 Pa

Change in volume of the gas ,ΔV= 1.15 m^3

Here the work is done by system so, the value of work done will be negative.

W=-P\Delta V

=W=-102000 Pa\times 1.15 m^3=117,300 J

The work done by the gas is 117,300 Joules.

8 0
4 years ago
In a __________ change the identity of the substance is
galben [10]

Answer:

physical - arrangement of molecules changes

chemical - identity changes

Explanation:

7 0
4 years ago
Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene
kenny6666 [7]

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}, 2-bromo-<em>2</em>-methylbutane;
  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}, 2-bromo-<em>1</em>-methylbutane.

It is expected that \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3} would end up being the dominant product.

Explanation

Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions \text{H}^{+} and negatively-charged bromide ions \text{Br}^{-}.

\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbo<em>cation</em> with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3};
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}.

The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has <em>three</em> carbon-carbon single bonds after the addition of the proton \text{H}^{+} as opposed to <em>two</em> as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.  

Both carbocations would then combine with bromide ions to produce a neutral halocarbon.  

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}

The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-<em>2</em>-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.

3 0
3 years ago
A metal atom and a non-metal atom bond together. What type of bond do they form?
Sliva [168]

Answer: Iconic

Explanation:Ionic bonds are formed through the exchange of valence electrons between atoms, typically a metal and a nonmetal.

6 0
3 years ago
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