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4 years ago

20 POINTS TO ANYONE WHO CAN ANSWER THIS QUESTION!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

2 answers:

Masteriza [31]4 years ago

8 0

Answer:

Sight, smell, sound, taste and touch, you have five senses. In your everyday lives, each of them is really important. Every moment of the day you use at least one of your five senses, including when you are sleeping! Together, the senses let your brain know what's happening around you.

~DjMia~

Grace [21]4 years ago

6 0

Answer:

touch, smell, hear, taste, and sight

Explanation:

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If i were to visit mystery island in january would i have to bring an umbrella?

Nitella [24]

Yes, you would have to bring an umbrella because if the island is named mystery you never know when it'll rain.

3 0

3 years ago

What’s is the percent composition by mass of oxygen in magnesium oxide , Mg0

Advocard [28]

Answer:

40%

Explanation:

We'll begin by obtaining the molar mass of MgO. This is illustrated below:

Molar Mass of MgO = 24 + 16 = 40g/mol

Observing the formula MgO, we have 1 atom of O in it.

The percentage composition by mass of oxygen in MgO is given by:

Mass of O/Molar Mass of MgO x 100

= 16/40 x 100 = 40%

6 0

3 years ago

Why do you need to remove carbon dioxide and water vapour to produce liquid air

AleksandrR [38]

Answer:

because these substances solidify when cooled and would clog the pipes of the air liquefaction plant.

7 0

3 years ago

Na2CO3 reacts with dil.HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl ,

puteri [66]

Answer:

See explanation

Explanation:

Equation of the reaction;

Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)

Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3

Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl

1 mole of Na2CO3 reacts with 2 moles of HCl

0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl

Hence Na2CO3 is the limiting reactant

Since there is 0.6 moles of HCl present, the number of moles of excess reagent=

0.6 moles - 0.4 moles = 0.2 moles of HCl

1 mole of Na2CO3 forms 1 mole of water

0.2 moles of Na2CO3 forms 0.2 moles of water

Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water

1 mole of Na2CO3 yields 1 mole of CO2

0.2 moles of Na2CO3 yields 0.2 moles of CO2

1 mole of CO2 occupies 22.4 L

0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP

Hence;

V1=4.48 L

T1 = 273 K

P1= 760 mmHg

T2 = 27°C + 273 = 300 K

P2 = 760 mmHg

V2 =

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 × 4.48 × 300/760 × 273

V2= 4.9 L

The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.

8 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

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

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago