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3 years ago

Please select the best answer and click "submit."

1 answer:

vladimir1956 [14]3 years ago

5 0

When it comes to physical changes like phase changes, there are two types of heat energy: sensible heat and latent heat. Sensible heat is the heat absorbed/released when you heat the substance but it doesn't change phase. An example would be heating lukewarm water. The substance is liquid all throughout. Latent heat, on the other hand, is the heat absorbed/released when there is a phase change. An example would be boiling water, because it changes liquid to vapor.

Hence, for freezing liquid, you use the latent heat, specifically the heat of fusion. The answer should be

2.5 g * (1 mol/18.02 g) * 6.03 kJ/mol = 0.84 kJ/mol

The answer is not in the choices. You only use Hvap if you boil water.

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A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

3 years ago

How many moles of carbonate are there in sodium carbonate

vaieri [72.5K]

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>

The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

= 46 + 12 + 48 = 106g/mol

mass of CO3 = 12 + 48 = 60g

no. of moles of CO3 = 60/106

no. of moles of CO3 = 0.566mol

Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

5 0

2 years ago

Read 2 more answers

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one ele

liubo4ka [24]

Answer:

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms? It is a negative ion that has one more valence electron than a neutral bromine atom.

Explanation:

7 0

2 years ago

How is the mass of 1 mole of an element determined

Tatiana [17]

One mole (abbreviated mol) is equal to 6.022×1023 molecular entities (Avogadro's number), and each element has a different molar mass depending on the weight of 6.022×1023 of its atoms (1 mole). The molar mass of any element can be determined by finding the atomic mass of the element on the periodic table.

3 0

3 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

3 years ago