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Elodia [21]
3 years ago
7

5. How is a cell wall different from a cell membrane?

Chemistry
1 answer:
Lina20 [59]3 years ago
5 0
Cell membrane is present is all cells while cell wall is only present in plants, bacteria, fungi and algae. ... Cell wall is made up of cellulose whereas cell membrane is made up of lipids and proteins.
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What is 25 °C in °F?
kifflom [539]

Answer:

77°F

Explanation:

(25°C × 9/5) + 32 = 77°F

7 0
3 years ago
1- Science that deals with the identification and quantification of the components of material systems
Elis [28]

Answer:

Analytical science

Explanation:

7 0
4 years ago
Most elements are ____________. The _____________ elements on the left of the stair-step line on the ______________ of Elements
Vlada [557]

Answer:

metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster  , malleable , ductile, opposite , shiny, moderately

Explanation:

metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster  , malleable , ductile, opposite , shiny, moderately Is the answer.

(Hope this helps can I pls have brainlist (crown)☺️)

4 0
3 years ago
Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylb
Vikki [24]

Answer:

None of these

Explanation:

Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.

Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :

Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.

The product formed is shown in mechanism does not mention in any of the options.

So, None of these is the answer

8 0
4 years ago
Read 2 more answers
The solubility product for barium sulfate is 1.1 × 10–10. calculate the molar solubility of barium sulfate.
Triss [41]
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,

 cC ⇔ aA + bB 

Ksp = {[A^a][B^b]}

Now, for the this problem, the reaction is

BaSO₄ ⇔ Ba²⁺ + SO₄²⁻

The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.

1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
8 0
3 years ago
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