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Alenkinab [10]
3 years ago
10

How much heat is required to melt 26.0 g of ice at its melting point?

Chemistry
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

Heat required to melt 26.0 g of ice at its melting point is 8.66 kJ.

Explanation:

Number of moles of water in 26 g of water: 26× \frac{1}{18.02} moles

                                                                      =1.44 moles

The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol.

we have relation as:

                                           q = n × ΔH

where:

q  = heat

n  = moles

Δ H  = enthalpy

So calculating we get,

                                        q= 1.44*6.02 kJ

                                        q= 8.66 kJ

We require 8.66 kJ of energy to melt 26g of ice.

                       

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Any system within the Earth system is considered an open system. Because energy flows freely into and out of systems, all systems respond to inputs and, as a result, have outputs.

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Consider the heating curve for water. A graph of the heating curve for water has time in minutes on the horizontal axis and Temp
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Answer:

0°C.

Explanation:

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In this case, given the heating curve of water on the attached document, we can notice that at 0 °C the solid starts melting, which means that the melting point is reached. Melting point is known as a physical change whereby a solid changes to liquid by the addition of heat as it allows the molecules to separate to each other.

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Differentiate between sol,aerosol and solid soluti​
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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

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7 0
2 years ago
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