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Gelneren [198K]
2 years ago
13

A 100 milliliters (mL) sample of a liquid is poured from a beaker into a graduated cylinder. Which of the following best explain

s what happened to the liquid?
Chemistry
1 answer:
olga_2 [115]2 years ago
7 0

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, when a liquid of a specific volume is transferred/poured from a container (in this case, beaker) to another container (in this case, graduated cylinder), the volume reduces (however insignificant), this is because parts of the water molecules will be "glued" to the interior body-part of the transferring container (beaker). Hence, the volume of liquid will be a little lesser than 100 mL when transferred into the graduated cylinder.

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No, They need something to hold on to, such as dirt
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When energy is added to a substance at constant volume and pressure, its temperature increases except ________.
Travka [436]

A. When the substance is in its gaseous state.

<u>Explanation:</u>

When a substance is expanding against its constant volume and pressure, its temperature increases except when the substance is in gaseous state and not in liquid or solid state. So the internal energy increase in the system not only increases and maintaining the volume and pressure of the system remains constant in its gaseous phase. In the first law of Thermodynamics, it is used specifically that to especially in the case of gaseous system.  

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7 0
3 years ago
A train travels at a speed of 80 mph. How far does the train travel in 2 h 42 min?<br> miles
VikaD [51]
2h 42m = (2+42/60)h = 2.7h
80mi/h x 2.7h = (80*2.7)mi = 216mi
4 0
3 years ago
One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc
Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

\text{The new partial pressure for }N_2 \ gas}

P_1V_1=P_2V_2

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar

\text{The new partial pressure for }Ar \ gas}

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar

Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = 2.225 \ Bar

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

2.1  \ bar = 2.07  \ atm \\ \\3.4 \  bar = 3.36 \  atm

For moles N₂:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}

n = 0.08297 \ mol  \ N_2

For moles of Ar:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}

n = 0.2172 \ mol  \ Ar

\mathtt{total \  moles = moles \ of \  N_2 + moles  \ of \ Ar}

=0.08297 mol + 0.2037 mol \\                   = 0.2867 mol gases

Finally;

The final pressure of the mixture is:

PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}

P = 2.217 atm

P ≅ 2.24 bar

7 0
3 years ago
2. Calculating density of an object with a regular shape
Tatiana [17]

D= 0.10125kg/m^3

EXPLAINATION
Density =mass/volume
D= 226.8/20 x 8 x 14
D=226.8/2240
D= 0.10125kg/m^3
Hope it was helpfull!!
6 0
3 years ago
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