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2 years ago

How does the written lewis structure for potassium chloride differ from that of hydrogen chloride?

Chemistry

1 answer:

labwork [276]2 years ago

5 0

Lewis structure of both is similar:

H : CI (hydrogen chloride)

K : CI (potassium chloride)

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How many moles of Cobalt are in 27.4 grams of cobalt

xeze [42]

0.4649331785818406 is what 27.4 grams is converted to! You're welcome!! :)

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2 years ago

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Suppose that magnesium would react exactly the same as copper in this experiment. how many grams of magnesium would have been us

Vikentia [17]

The solution for this problem would be:

We are looking for the grams of magnesium that would have been used in the reaction if one gram of silver were created. The computation would be:

1 g Ag (1 mol Mg) (24.31 g/mol) / (2mol Ag)(107.87g/mol) = 0.1127 grams of Magnesium

6 0

2 years ago

An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an

timofeeve [1]

Answer:

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance: $M_{1} = 175 + 11M_{2}$

Water: $0.804M_{1} = 175 + 0.6M_{2}$

Using substitution method, we have:

$M_{1}$ = 220.77 kg/s

$M_{2}$ = 4.16 kg/s

The maximum production rate of solid crystal is $10M_{2}$ = 10*4.16 = 416 kg/s

Around evaporator:

$0.45M_{5} = 175$

$M_{5} = 175/0.45 = 389$ kg/s

Around the mixing point:

$M_{1} + M_{3} = M_{4} + M_{5}$

Solid crystal: $0.196M_{1} + 0.4M_{3} = M_{4}$

Using the last two equations, we can obtain:

$0.804M_{1} + 0.6M_{3} = M_{5}$

$0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5$

$M_{3} = 211.5/0.6 = 352.5$ kg/s

The rate of supplying fresh feed to obtain the production rate is:

$\frac{M_{3}}{M_{1}}$ = 352.5/220.77 = 1.6

7 0

3 years ago

What is the volume (in ml) of a 12.9 g piece of metal with a density of 7.25 g/cm3?

Leto [7]

Hey there:

1 cm³ = 1 mL

D = m / V

7.25 = 12.9 / V

V = 12.9 / 7.25

V = 1.779 cm³

6 0

2 years ago

Read 2 more answers

Use the following half-reactions to construct a voltaic cell:

velikii [3]

Answer: The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

Explanation:

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

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:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

4 0

2 years ago