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KIM [24]
4 years ago
13

When 13.6 grams of kcl are dissolved in enough water to create a 180-gram solution, what is the solution's concentration, expres

sed as a percent by mass? 4.6% kcl 6.8% kcl 7.6% kcl 8.2% kcl?
Chemistry
2 answers:
bonufazy [111]4 years ago
8 0
7.6% KCl I had the same problem on my semester exam for summer school
jek_recluse [69]4 years ago
6 0

Answer: 7.6% KCl

Explanation: Mass of solute is given as 13.6 grams and the mass of solution is 180 grams. The question asks to calculate the solution concentration, expressed as a percent by mass.

The formula for percent by mass is:

% by mass = \frac{solute}{solution}*100

Let's plug in the values in the formula and do the calculations.

% by mass = \frac{13.6g}{180g}*100  = 7.6%

So, the correct choice is C) 7.6% KCl.

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It's a mixture of H2O + a soluble substance like, salt.
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3 years ago
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For a particular redox reaction, Cr is oxidized to CrO 2 − 4 and Ag + is reduced to Ag . Complete and balance the equation for t
Sunny_sXe [5.5K]

Answer:

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Explanation:

We can balance the redox reaction of Cr and Ag⁺, in terms of two half-reactions, one for Ag⁺ and other for Cr:

Ag⁺   →   Ag      

In the above equation we need to balance the number of electrons, we know that the Ag⁺ is being reduced to Ag, so the reaction is:

Ag⁺ + e⁻ →  Ag   (1)

Now, we need to balance the half-reaction of Cr:

Cr   →  CrO₄²⁻  

From above, we know that the Cr is being oxidated to CrO₄²⁻, so we need to balance the number of electrons and the number of oxygen atoms. The Cr⁰ is being oxidated to Cr⁶⁺, so for the electron balance, we need to add 6e⁻ to the right side of the equation. Since the reaction is in a basic medium, the oxygen atoms will be balanced with OH⁻ ions as follows:          

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻  

The hydrogen atoms will be balanced using H₂O molecules:  

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻ + H₂O    

The balanced equation is:

Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O   (2)

Since the reaction (1) involves 1 electron and the reaction (2) involves 6 electrons, by increasing the reaction (1) six times and by the addition of the two reactions (1 and 2) we can have the net redox reaction:

6*(Ag⁺ + e⁻ →  Ag)  

<u>Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O</u>

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O                  

Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.

I hope it helps you!

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4 years ago
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Artyom0805 [142]
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Explanation:
Catalysts make such a breaking and rebuilding happen more efficiently. They do this by lowering the activation energy for the chemical reaction. Activation energy is the amount of energy needed to allow the chemical reaction to occur. The catalyst just changes the path to the new chemical partnership.

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Comment above 100 percent correct
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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
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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.

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3 years ago
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