Answer:- Third choice is correct, 17.6 moles
Solution:- The given balanced equation is:
Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4
We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.
From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.
It is a simple mole to mole conversion problem. We solve it using dimensional set up as:
2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})
= 17.6 mol KOH
So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.
Answer:
3.59x10^21 molecules
Explanation:
1mole of a substance contains 6.02x10^23 molecules.
Therefore, 1mole of C8H18 will also contain 6.02x10^23 molecules
1mole of C8H18 = (12x8) +(18x1) = 96 + 18 = 114g.
1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.
Therefore, 0.68g of C8H18 will contain = (0.68 x 6.02x10^23)/114 = 3.59x10^21 molecules
Answer:
Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).
Explanation:
The following reaction gives a product with the molecular formula C₄H₈O₂. The diagram of the structure of the product can be seen in the image attached below.
The reaction between C₂H₂(ONa)₂ and C₂H₄Br results in the formation of the product C₄H₈O₂ and 2NaBr.
This reaction undergoes an SN₂ mechanism since there is no stable carbocation formed. In the reaction -O⁻Na⁺ attacks the ortho position in C₂H₄Br to form C₄H₈O₂.
In SN₂ mechanism is a nucleophilic substitution reaction where one bond is formed while another one is broken simultaneously.
The mechanism for the reaction can be seen in the image attached below.
Learn more about nucleophilic substitution reaction here:
brainly.com/question/4699407?referrer=searchResults
Answer:
a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂
b) Ni(OH)₂
c) KOH
d) 0.927 g
e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M
Explanation:
a) The equation is:
2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)
b) The precipitate formed is Ni(OH)₂
c) The limiting reactant is:


From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:
Hence, the limiting reactant is KOH.
d) The mass of the precipitate formed is:
e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:


I hope it helps you!