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3 years ago

HELP 40 POINTS

Chemistry

2 answers:

padilas [110]3 years ago

8 0

Answer:

Explanation:

F2 is the only element with only 1 element and eveything else is formed by 2 non-metals

goldfiish [28.3K]3 years ago

3 0

Answer:d I believe

Explanation:

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The diagram shows four different locations in an atom. A shaded circle is shown, labeled nucleus. Two small points labeled one a

pshichka [43]

<h2>Answer : 2</h2><h3>Explanation :</h3>

The shaded area labeled as 2 is likely to have subatomic particle that are constantly in motion.

The subatomic particles of the atom that are constantly in motion are called as electrons. The probability of finding electrons in the region shaded and labeled as 2. Electrons are in continuous motion in orbits around the nucleus of the atom. Therefore, the answer is 2.

3 0

3 years ago

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What are the phase change processes taking place when a substance is in equilibrium between liquid and gas phases? A. Melting an

lawyer [7]

C. <span>Evaporating and condensing. When the liquid becomes gas is called evaporation. When gas become liquid it's condensation. </span>

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3 years ago

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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as 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:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

Only answer if you know please thank you

3241004551 [841]

Explanation:

1. liquid

2.celsius

3.true

4.evaporation since molecule releasing energy

5. this graph only shows exothermic phase change

7 0

3 years ago

4. The solubility of acetanilide in hot water is 5.5 g / 100 mL at 100oC and its solubility in cold water is 0.53 g / 100 mL at

Triss [41]

Answer:

89,4%

Explanation:

If you have a solutio of 2,5g of acetanilide in 50mL of water and you warm this solution to 100°C you will dissolve all acetanilide because the maximum solubility in 50mL will be:

5,5g / 100mL → 2,75g / 50mL.

Then, if you cold the water to 0°C the solubility in 50mL will be:

0,53g / 100mL → 0,265g / 50mL.

That means you will precipitate:

2,5g - 0,265g = <em>2,235g of acetanilide</em>

The theoretical percent recovery will be:

2,2365g / 2,5g ×100 = <em>89,4%</em>

I hope it helps!

7 0

3 years ago