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3 years ago

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Initially, 0.740 mol of A is present in a 5.00 L solution. 2 A ( aq ) − ⇀ ↽ − 2 B ( aq ) + C ( aq ) At equilibrium, 0.200 mol of

C is present. Calculate K.

1 answer:

drek231 [11]3 years ago

4 0

Answer:

POINTZZZ

Explanation:

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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

How many atoms in 1kg of platinum<br> a 2.5x10^24<br><br> b 3.1x10^24

Nutka1998 [239]

Answer:

3.1x10^24 it will be in 1 kg of platinum

6 0

3 years ago

1. When this equation is balanced, Fe + 02 -.--> Fe203, what is the

Vilka [71]

Fe + O2 → Fe2O3

After balancing the eq.

4Fe + 3O2 → 2Fe2O3

Hope this will help u mate :)

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3 years ago

For the following species, draw a Lewis structure for one important resonance form. Include all lone pair electrons and assign f

Gemiola [76]

Answer:

The Lewis structures are shown in the figure

Explanation:

The lewis structure will be drawn using following steps

i) we will calculate the total number of valence electrons in the molecule

ii) will assign one bond (two electrons in between two atoms)

iii) then distribute the rest of the valence electrons as lone pair or shared pair based on completion of octet.

The structure of each molecule is given in the figure.

Valence electrons:

$NO_{2}^{+}$

V.E = 5 +(2)6-1=16

$NO_{2}F$

V.E =5 +(2)6+7=24

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3 years ago

Freezing and thawing water can cause a rock to break down into smaller pieces. Which part of the rock cycle is this an example o

zloy xaker [14]

Answer:

Deposition

Explanation:

It’s breaking down the rock :]

4 0

3 years ago

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