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3 years ago

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Balance and classify the following equations as synthesis, decomposition, single replacement, double replacement, or combustion:

a. C4H10 + 02 -> CO2 + H20 Reaction Type: b. ___ HNO3 + Mg(OH)2 -> Mg(NO3)2 + H20 Reaction Tyne.

1 answer:

vodomira [7]3 years ago

8 0

See which one is the bigger equitation to classify the smallest in order

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An aqueous solution contains 32.7% kcl (weight/weight %). how many grams of water are contained in 100 g of this solution

igor_vitrenko [27]

Weight/weight percentage is the percentage of mass of solute from the mass of the whole solution
w/w % = 32.7 %
this means that in a solution of 100 g - mass of KCl is 32.7 g
since solution is made of water and KCl
then the mass of water is - 100 - 32.7 = 67.3 g
therefore in 100 g of solution - 67.3 g of the solution is water

5 0

3 years ago

Please help me with this.<br><br><br>And show all work as well ASAP!!

qaws [65]

Answer: The partial pressure of oxygen is 187 torr.

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

$p_1=x_1p_{total}$

where, x = mole fraction

$p_{total}$ = total pressure

$x_{oxygen}=\frac{\text {moles of oxygen}}{\text {total moles}}$ ,

$x_{oxygen}=\frac{3.0}{12.33}=0.243$ ,

$p_{oxygen}=0.243\times 770torr=187torr$

Thus the partial pressure of oxygen is 187 torr.

6 0

3 years ago

1.1 Outline a method for separating the chalk from potassium chloride,

miss Akunina [59]

Answer:

Explanation:

To separate the a mixture of chalk and potassium chloride, we must not that chalk is calcium carbonate compound, CaCO₃.

Calcium carbonate has low solubility in water. KCl is readily soluble in water and it is also an ionic compound.

To separate a mixture of compounds with various solubility, we can carryout dissolution, filtration and evaporation.

We first pour pure water into the mixture. Water will dissolve the potassium chloride readily.

Then using a filter paper we filter out the suspended chalk particles. Leave the filtrate to then dry and collect it.

The solution filtered should be evaporated to dryness. This will leave the KCl behind from the solution.

5 0

3 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

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

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

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:

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

<u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

<u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

<u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

<u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

<u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

3 years ago

Describe the heat involved if you touch a flask containing an on-going endothermic reaction

juin [17]

The flask would feel cooler than when the reaction first started.

8 0

4 years ago