In which liquid does solid brass float?

Chemistry

2 answers:

zimovet [89]4 years ago

7 0

It’s A , Liquid Mercury

ch4aika [34]4 years ago

6 0

Answer:

Explanation:

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What mass in grams of KCl (74.55 g/mol) is produced from 65.8 grams of KClO3 (122.55 g/mol) if the reaction has a yield of 67.2%

marishachu [46]

Y’all what is the answer

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3 years ago

A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1

icang [17]

Answer:

Approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

$\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C$ .

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took $2911\; \rm J$ of energy to raise the temperature of this sample by $\Delta T = 92.6\; \rm ^\circ\! C$ . Therefore, raising the temperature of this sample by $1\; \rm ^\circ\! C$ (unit temperature) would take only $\displaystyle \frac{1}{92.6}$ as much energy. That corresponds to approximately $31.436\; \rm J$ of energy.

On the other hand, the energy required to raise the temperature of this material by $1\; \rm ^\circ\! C$ is proportional to the mass of the sample (also assuming no phase change.)

It took approximately $31.436\; \rm J$ of energy to raise the temperature of $57.07\; \rm g$ of this material by $1\; \rm ^\circ C$ . Therefore, it would take only $\displaystyle \frac{1}{57.07}$ as much energy to raise the temperature of $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C\!$ . That corresponds to approximately $0.551\; \rm J$ of energy.

In other words, it takes approximately $0.551\; \rm J$ to raise $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C$ . Therefore, by definition, the specific heat of this material would be approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .