Ne mutlu TÜRKÜM DİYENE...

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]

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

Explanation:

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:

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

Calculating the freezing point:

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

Calculating the boiling point:

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,

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