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s344n2d4d5 [400]
2 years ago
10

A 2.0-g sample of washing soda, Na CO • 10H O, has carbon atoms. How many oxygen atoms are present in 2.0g of washing soda?

Chemistry
1 answer:
vovikov84 [41]2 years ago
7 0

Description is on the picture in attachment

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Describe the difference in structure between beH2 and CaH2
katen-ka-za [31]
Hello!

BeH₂ is a linear molecule, while CaH₂ is an angular molecule.

The difference between these two molecules is given by the number of electrons they have. Be is in the 2nd period of the Periodic table, and the ion Be⁺² doesn't have any free electron pairs when bonding to H. Ca is in the 4 period of the periodic table, meaning that it has more electrons, and the ion Ca⁺² has two free electron pairs when bonding to H that makes the molecule angular by pushing the bonds at an angle by sterical hindrance.  

Have a nice day!
7 0
2 years ago
If charges flow very slowly through a metal, why does it not require several hours for a light to come on when you throw a switc
Harlamova29_29 [7]
T<span>he charges themselves may not move fast, but the force upon them does. The electric field set up by the battery or generator propagates through the wires at the speed of light.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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6 0
2 years ago
A beaker has 0.2 M of Na2SO4. What will be the concentration of sodium and sulfate ions?
netineya [11]

The concentration of sodium and sulphate ions are [Na^+] = 0.4 M, [SO_4^2-] = 0.2 M

Explanation:

The molar concentration is defined as the number of moles of a molecule or an ion in 1 liter of a solution.

In the given solution, the concentration of the salt sodium sulphate is 0.2M. So, 0.2 moles of sodium sulphate is present in 1 liter of solution.

Assuming 100% dissociation,

1 molecule of sodium sulphate gives 2 ions of sodium and 1 ion of sulphate.

So 0.2 moles of sodium sulphate will give 0.4 moles of sodium ions and 0.2 moles of sulphate ions.

7 0
2 years ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):

I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).

See figure 1

I hope it helps!

7 0
1 year ago
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