Whats the atomic radius of tin (Sn)

The reaction below is at equilibrium. What would happen if more carbon were

schepotkina [342]

Explanation:

$C_{(s)} + O_{2(g)} \rightleftharpoons CO_{2(g)}$

According to le ch-atelier's principle when a chemical reaction at equilibrium subjected to any stress, then the equilibrium shifts in that direction in which effect of stress is reduced.

When we would add more number of carbon in above reaction then rate of forward reaction will increase to attain the equilibrium again.

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5 0

2 years ago

How can one kg of iron melt more ice than 1 kg lead at 100 °C

Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{iron}$ = Heat obtained from the iron by the ice

ΔQ $_{iron}$ = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{lead}$ = Heat obtained from the iron by the ice

ΔQ $_{lead}$ = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0

3 years ago

Explain why vanadium (radius=134 pm) and copper (radius=128 pm) have nearly identical atomic radii, even though the atomic numbe

Viktor [21]

There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.

Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons; density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.

7 0

3 years ago

14. What is a major disadvantage of using solar power?

sleet_krkn [62]

I believe that number 27 is A

8 0

3 years ago

Read 2 more answers

What is the balanced form of the following equation?. Br2 + S2O32– + H2O → Br1– + SO42– + H+.

zloy xaker [14]

For the equation to be balanced, the Atom's coefficient on the left side and the right side of the equation has to be equal

so, the answer would be :

Br2 + S2032- + 5H20 -- > BR2- + 2S02- + H+

Hope this helps

8 0

3 years ago