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Gemiola [76]
3 years ago
12

To what volume will a 2.33 l sample of gas expand if it is heated from 30.0 ∘c to 300.0 ∘c?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0
According to Charle's Law,
V/T =  constant
where 
V = volume
T =  absolute temperature.

The law holds when the pressure of an ideal gas is held constant while temperature changes.

V₁ = 2.33 L, initial volume
T₁ = 30°C = 30+273 K = 303 K, initial temperature
T₂ = 300°C = 300+273 K = 573 K, final temperature

The final volume, V₂, is given by
V₂/T₂ = V₁/T₁
V₂ = (573 K)*(2.33/303  L/K)
     =  4.406 L

Answer: 4.41 L

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What is the density of an object with a mass of 140 g and a volume<br> of 6 ml
Strike441 [17]

Answer:

<h2>23.33 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{140}{6} =  \frac{70}{3}   \\  = 23.3333...

We have the final answer as

<h3>23.33 g/mL</h3>

Hope this helps you

5 0
3 years ago
Any mixture that is heterogeneous on a microscopic level is a
Nadya [2.5K]
Solution is the answer.
5 0
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Why does a permeable rock weather chemically at a fast rate?
ki77a [65]
<span>Water soaks into it easily and during freeze-thaw cycles would be more likely to break apart because of the water freezing and thawing inside the rock. The water freezes and expands inside the rock, causing the surface to break off in flakes from the pressure of the expanding ice inside.</span>
3 0
3 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
Please help I will mark brainly
wolverine [178]
The first one i think
8 0
3 years ago
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