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3 years ago

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Max discovered a beaker full of transparent liquid in the science lab. He hypothesized that since the liquid is transparent like

water, it is water. He shared this conclusion with his friends. His friends questioned his conclusion, but Max had no evidence to back his thoughts. What step did Max miss in his scientific methods?
A.
Make an initial observation.
B.
Test a hypothesis by gathering data.
C.
Form a conclusion.
D.
State a hypothesis.

1 answer:

S_A_V [24]3 years ago

6 0

B. Test a hypothesis by gathering data

Seriously though, what type of guy thinks "oh hey there's a transperent liquid in the science lab? yeah it's water."

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Answer:

q = 38,5 kJ

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q = 4,184J/g°C×92,0g×100°C

q = 38493 J, in kilojoules:

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<em></em>

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Answer:

29273.178 joules have been added to the gas for the entire process.

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Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

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2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

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$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) <em>Bob further heats the gas at constant pressure until its volume is doubled</em>:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

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