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3 years ago

Use the graph to answer both parts of this question about these three minerals. Do not use calculations.

Chemistry

2 answers:

marissa [1.9K]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf Mercury \ is \ the most \ dense \ and \ titanium \ is \ the \ least \ dense}}$

Explanation:

The formula for density is:

$d=\frac{m}{v}$

As mass and volume increase, so does the density.

In this problem, the density is the line. So, the steeper the slope of the line, the greater the density.

Mercury (Hg) is the most dense because it has the steepest slope.
Copper (Cu) is more dense than titanium but less dense than mercury.
Titanium (Ti) is the least dense because it's slope is the least steep.

So, mercury is the most dense and titanium is the least dense.

11111nata11111 [884]3 years ago

6 0

Answer:

Mercury is the most dense whereas Titanium is the least dense.

Explanation:

Since Density = Mass / Volume, the slope of the line can immediately tell us its relative density between the metals.

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a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

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

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

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3 years ago

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marusya05 [52]

The answer should be d

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baherus [9]

The answer is 7 !! :)

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2 years ago

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dimaraw [331]

Answer:

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Alina [70]

Answer:

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