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monitta
3 years ago
6

What quantity of energy, in joules, is required to raise the temperature of 425 g of tin from room temperature, 25.0 °C, to its

melting point, 231.9 C, and then melt the tin at that temperature? (The specific heat capacity of tin is 0.227 J/g K, and the heat of fusion of this metal is 59.2 Jig) Energy required
Chemistry
1 answer:
Rama09 [41]3 years ago
6 0

Answer:

Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is 45.249 kiloJoules.

Explanation:

Mass of the tin ,m= 425 g

Heat capacity of the tin ,c= 0.227 J/g K

Initial temperature of the tin ,T_1= 25.0 °C = 298.15 K

Final temperature of the tin, T_2= 231.9 °C = 505.05 K

Let the heat required to change the temperature of tin from 298.15 K to 505.05 K be Q.

Q=mc\times (T_2-T_1)

=425 g\times 0.227 J/g K\times (505.05K - 298.15 K)=19,960.68 J=19.961 kJ

Heat required to melt tin at 505.05 K be Q'

The heat of fusion of tin metal =\Delta H_{fus}=59.2 J/g

Q'= m\times \Delta H_{fus}=425 g\times 59.2 J/g=25,287.5 J=25.288 kJ

Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is:

= Q+Q' =  19.961 kJ + 25.288 kJ = 45.249 kJ

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How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)
nikdorinn [45]

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

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H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200.  * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.

8 0
3 years ago
rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first
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The chemical equation for the protonation of carbonate ion in the presence of water follows:

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Explanation:

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3 years ago
Read 2 more answers
A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L
PtichkaEL [24]

Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

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2) Formula:

Used combined law of gases:

  PV / T = constant

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3) Solution:

Solve the equation for V₂:

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Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

You can learn more about gas law problems reading this other answer on

Explanation:

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