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labwork [276]
3 years ago
13

The temperature inside your freezer is 0 degrees Celsius. You place a balloon with an initial temperature of 30 degrees C and a

volume of 0.75 liters in the freezer. Once the balloon is fully chilled, what will its volume be?
Remember to pay close attention to the units of temperature before beginning your calculations.
Chemistry
1 answer:
blsea [12.9K]3 years ago
4 0

Answer:

V=0.68L

Explanation:

For this question we can use

V1/T1 = V2/T2

where

V1 (initial volume )= 0.75 L

T1 (initial temperature in Kelvin)= 303.15

V2( final volume)= ?

T2 (final temperature in Kelvin)= 273.15

Now we must rearrange the equation to make V2 the subject

V2= (V1/T1) ×T2

V2=(0.75/303.15) ×273.15

V2=0.67577931717

V2= 0.68L

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How many liters are equivalent to 43 milliliters?
marysya [2.9K]

Answer:

The answer is B) 0.043 L

Explanation:

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8 0
3 years ago
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Rank the following steps involved in preparing a dilute solution from a stock solution in order from first to last.
Serggg [28]

Answer: C - B - A - D

Explanation:

-Obtain a flask that is thoroughly cleaned

-Add the desired amount of stock solution

-Add deionized water until the volume reaches the mark on the flask.

- Cap the flask and invert several times to thoroughly mix the solution.

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3 years ago
List 10 things that need a pull to move.
LUCKY_DIMON [66]

Answer:

a cabinet

a door

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4 0
3 years ago
For the process 2SO2(g) + O2(g) --> 2SO3(g),
CaHeK987 [17]

Answer:

–187.9 J/K

Explanation:

The equation that relates the three quantities is:

\Delta G = \Delta H - T \Delta S

where

\Delta G is the Gibbs free energy

\Delta H is the change in enthalpy of the reaction

T is the absolute temperature

\Delta S is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K

8 0
4 years ago
Suppose you were calibrating a 100.0 ml volumetric flask using distilled water. the flask temperature was at 20°c, and you assum
kicyunya [14]
The density of water at 12°C is lower than the density of water at 20°C.

Now density is related to volume as per: density = mass / volume =>

mass = volume * density.

So, the greaer the density the higher the mass of the same volume of water.

Therefore, 100.0 ml of water at 12°C has a mass greater than 100.00 ml of water at 20 °C.
4 0
4 years ago
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