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3 years ago

What is the mass of one mole of calcium carbonate

Chemistry

2 answers:

vlada-n [284]3 years ago

6 0

100.08 grams
Use train track method!

valkas [14]3 years ago

3 0

The mass of one mole of calcium carbonate should be 100.087g. Hope this helps!

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Solid solutions that are mixtures of elements with metallic properties are known as_____

marshall27 [118]

Alloys. hope this helps.

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3 years ago

The pollen grains move in water randomly because of

polet [3.4K]

A the collisions between water particles and polllen grains

6 0

3 years ago

18.

Leno4ka [110]

Answer:

It helps the body remove heat through sweating.

Explanation:

When the weather is hot, the body tries to keep cool by sweating. The high specific heat capacity means that the body doesn't have to lose much water to stay cool.

The high specific heat capacity of water doesn’t heat the body, but it slows down the rate of heat loss when the weather is cool.

B is wrong. The body uses glucose, not water, as an energy source.

C is wrong. The high specific heat capacity of water is not connected with the body's ability to store it.

D is wrong. The high specific heat capacity of water doesn't heat the body, but it slows the rate at which it cools.

6 0

3 years ago

Read 2 more answers

The speed of light is around 6.706×10^8 miles per hour. What is the speed of light in units of miles per minute?

Sindrei [870]

Answer: The speed of light in miles per minutes is $1.117\times 10^7miles/min$

Explanation:

We are given the speed of light is $6.706\times 10^8miles/hr$ and we need to convert it into miles/min. So, we use the converion factor:

1 hour = 60 minutes

Converting that quantity into miles/minutes, we get:

$(\frac{6.706miles}{1hr})(\frac{1hr}{60min})=\frac{1.117\times 10^7miles}{min}$

Hence, the speed of light in miles per minutes is $1.117\times 10^7miles/min$

7 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