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Neporo4naja [7]
3 years ago
14

How do decantation and filtration differ which should be faster

Chemistry
1 answer:
ahrayia [7]3 years ago
8 0
Decantation describes when a substance settles to the bottom, think of the Hershey syrup and milk, the syrup slowly settles to the bottom. Filtration describes a mixture and as an example Sand and Water and separation comes from setting up a filter paper. Filtration is usually faster.
You might be interested in
What is number “4” in SiCi4?
JulsSmile [24]

It is a subscript

Also do you mean Cl not Ci. I assume you do because then it means 4 chlorine in that compound.

5 0
4 years ago
(b) what is the major product of the reaction at very low temperatures?
svlad2 [7]
The reaction is low b/c of  the tempature. that's why.
8 0
3 years ago
Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr
leonid [27]
The law that states the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pressure conditions would be the Charles Law. It <span>is an experimental gas </span>law<span> that describes how gases tend to expand when heated. Hope this answers the question.</span>
6 0
4 years ago
How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

6 0
3 years ago
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

4 0
3 years ago
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