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Sav [38]
3 years ago
9

You need a 20% alcohol solution. on hand, you have a 240 ml of a 15% alcohol mixture. you also have 40% alcohol mixture. how muc

h of the 40% mixture will you need to add to obtain the desired solution?
Chemistry
1 answer:
juin [17]3 years ago
4 0
0.15*240=36 ml of alcohol in <span>240 ml of a 15% alcohol mixture
0.4x = </span>ml of alcohol in x ml of a 40% alcohol mixture
0.2(x+240)= ml of alcohol in (x+240) ml of a 20% alcohol mixture

0.15*240 + 0.4x = 0.2(x+240)
36+0.4x=0.2x+48
0.2x = 12
x=12/0.2=120/6=20 ml  of a 40% alcohol mixture
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In the reaction FeCl2 + 2NaOH -&gt;Fe(OH)2(s) + 2NaCl, if 6 moles of FeCl2 are added to 6 moles of Na0H, how many moles of NaOH
pogonyaev

Answer : The correct option is, (B) 6 mole

Explanation :

Given moles of FeCl_2 = 6 moles

Given moles of NaOH = 6 moles

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

FeCl_2+2NaOH\rightarrow Fe(OH)_2+2NaCl

From the given balanced reaction, we conclude that

As, 1 moles of FeCl_2 react with 2 moles of NaOH

So, 6 moles of FeCl_2 react with \frac{2}{1}\times 6=12 moles of NaOH

From this we conclude that, FeCl_2 is an excess reagent and NaOH is a limiting reagent because the given moles are less than the required moles and it limits the formation of product.

Thus, the number of moles of NaOH used up in the reaction = Required moles of NaOH - Given moles of NaOH

The number of moles of NaOH used up in the reaction = 12 - 6 = 6 moles

Therefore, the number of moles of NaOH used up in the reaction will be, 60 moles

4 0
3 years ago
Read 2 more answers
Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
A scientist needs to collect a 0.050 mole sample of helium, but needs to know how large his helium container should be. What vol
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We can use the ideal gas equation:

PV = nRT

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T = 400K

We will have to convert from Pa to atm or viceversa.

101325 Pa________1 atm

202600 Pa________x = 2.00 atm

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V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm = 0.82 liters = 820 mililiters



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Answer:

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Explanation:

calculation

6 0
2 years ago
Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass
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Answer:

1.8g

Explanation:

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= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

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