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ahrayia [7]
3 years ago
15

Ne mutlu TÜRKÜM DİYENE...

Chemistry
2 answers:
Nata [24]3 years ago
7 0

what is the question?


alexandr1967 [171]3 years ago
6 0
English please 
and questions
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In order to become stable, Potassium, K, will _______ and have a resulting charge of _____.
svlad2 [7]

Answer:

D

Explanation:

it doesn't matter how many protons k has all the answers have electron with the positive charge when is negative the question is trying to distract you D is the only one that would use the electrons sign correctly

4 0
2 years ago
What role does the wire play in voltaic cell
VashaNatasha [74]
It allows electrons to flow from the anode to the cathode.
8 0
2 years ago
What makes the outer shell electrons important is that they
klemol [59]
A, because the number of valence shell electrons (outer shell electrons) tells us how much the element or compound wants to bond or give up electrons. Most compounds and elements want to have eight valence ectrons in it's outer ring. So if an atom is far away from having eight, it will want to react more often.
3 0
2 years ago
Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a
fiasKO [112]

Answer:

                    SEPARATION SCHEME FOR  CATIONS

GIVEN  CATIONS : Ag^{+} \ ,  Fe^{3+} , Cu^{2+}, Ni^{2+}

     

    Step 1:   Add 6mol/dm^3 of HCl to the mixture solution

    Result : This would cause a precipitate of AgCl to be formed

    Reaction :  Ag^{+} _{(aq)} + Cl^{-} _{(aq)}  ---------> AgCl(ppt)

    Step 2 : Next is to remove the precipitate and add H_2S to the remaining          

                 solution in the presence of 0.2 \ mol/dm^3 of HCl

     Result : This would cause a precipitate of CuS to be formed

     Reaction :  Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)

 

     Step 3: Next remove the precipitate then add 6 \ mol/dm^3 of aqueous      

                 NH_3 (NH_3 \cdot H_2 O) , process the solution in a centrifuge,when the  

                 process  is done then sort out the  precipitate from the  solution

                 Now this precipitate is   Fe(OH)_3 and the remaining solution

                contains  (Ni (NH_3)_6)

                 Next take out the precipitate to a different beaker and add HCl

                to it   this will dissolve it, then add a drop of NH_4SCN this will

                form  a precipitate  Fe(SCN)_{6}^{3-} which will have the color of

                 blood  indicating the presence of Fe^{3+}

             

   Reaction :   F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}

                        Fe (OH)_{(s)} _3  + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}

                         Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}

                      Now the remaining mixture contains Ni^{2+}

     

       

Explanation:

6 0
3 years ago
Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol
pishuonlain [190]

<u>Answer:</u> The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

<u>Explanation:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g

  • <u>Calculating the freezing point:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.99°C/m

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62 g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC

Hence, the freezing point of solution is -117.54°C

  • <u>Calculating the boiling point:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_b = molal boiling point elevation constant = 1.20°C/m.g

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62  g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC

Hence, the boiling point of solution is 80.48°C

3 0
3 years ago
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