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VladimirAG [237]
3 years ago
13

The density of a sample of metal is calculated using these three different sets of data: 1.2 g/mL, 1.4 g/mL, and 1.1 g/mL. If th

e true density of the metal is 1.2 g/mL, what can be said about these calculated density values? Are they accurate and precise, just one, or neither?
Chemistry
1 answer:
Iteru [2.4K]3 years ago
8 0
Accurate data means the data experimentally obtained are close to the true value. Precise data means the data obtained are close to one another. In this case, the data are close to the true value which is 1.2 and the data are relatively close to one another. Hence the set is both accurate and precise.
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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
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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}

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Please find the definition and further explanation below

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Decomposition its the answer
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