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Blizzard
3 hours ago
4

What is the molality of a solution containing 0.25 g

Chemistry
You might be interested in
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
eduard [2782]

Answer:- 64015 J

Solution: The calorimeter contains 4250 mL of water, which is at a temperature of 22.55 degrees Celsius.

The water's density is 1 gram per mL.

Thus, the mass of water = 4250mL(\frac{1g}{1mL}) = 4250 grams.

After introducing the hot copper bar, the final temperature of the water reaches 26.15 degrees Celsius.

Thus, \Delta T for the water = 26.15 - 22.55 = 3.60 degrees Celsius.

The specific heat capacity of water is 4.184 \frac{J}{g.^0C}.

To determine the heat absorbed by the water, we can use the following formula:

q=mc\Delta T

where q represents heat energy, m refers to mass, and c indicates specific heat.

Now let's substitute the values into the equation to perform the calculations:

q=4250g*\frac{4.184J}{g.^0C}*3.60^0C

q = 64015 J

Therefore, the water absorbs 64015 J of heat.



5 0
3 months ago
Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of
castortr0y [3046]
The temperature difference after 3 hours is 5.16 K. Given that the moles of O₂ inhaled rate at 0.02 mole/min, which converts to 1.2 mole/hour, we know the average heat released during metabolism is 7.2 kJ/h·kg. Therefore, the amount of heat generated within 3 hours will be 7.2 kJ/h·kg multiplied by 3 hours, giving a result of 21.6 kJ/kg, or 21.6 x 10³ J/kg. Applying the formula Qp = Cp x ΔT, and taking the body's heat capacity to be 4.18 J/g·K, we find ΔT = 5.16 K.
6 0
2 months ago
the image above shows a chamber with a fixed volume filled with gas at a pressure of 1560 mmHg and a temperature of 445.0 K. If
VMariaS [2998]

Answer:

The new gas pressure within the chamber registers at 1,093.75 mmHg

Explanation:

The Gay-Lussac Law establishes a relationship between a gas's pressure and temperature when volume remains constant. This principle asserts that gas pressure is directly tied to its temperature: as temperature increases, pressure rises, and conversely, as temperature falls, pressure also diminishes. Therefore, the Gay-Lussac law can be depicted mathematically as:

\frac{P}{T} =k

Given an initial and final state of gas, we can apply the following formula:

\frac{P1}{T1} =\frac{P2}{T2}

In this scenario:

  • P1= 1560 mmHg
  • T1= 445 K
  • P2=?
  • T2= 312 K
<psubstituting:>

\frac{1560 mmHg}{445 K} =\frac{P2}{312 K}

Calculating:

P2=\frac{1560 mmHg}{445 K} *312K

P2=1,093.75 mmHg

The new gas pressure inside the chamber is 1,093.75 mmHg

</psubstituting:>
7 0
3 months ago
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