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enyata [817]
3 years ago
15

Which has a greater volume, 10 grams of water or 10 grams of acetone?

Chemistry
1 answer:
WINSTONCH [101]3 years ago
6 0

Density is defined as the ratio of mass to the volume.

Density = \frac{Mass}{Volume}          (1)

Mass of water = 10 grams

Mass of acetone  = 10 grams

Density of water  = 1 \frac{g}{cm^{3}}

Density of acetone  = 0.7857 \frac{g}{cm^{3}}

Put the value of density of water and its mass in equation (1)

1 \frac{g}{cm^{3}} =  \frac{10 g}{volume}

Volume of water =  10 cm^{3}

Put the value of density of acetone and its mass in equation (1)

0.7857 \frac{g}{cm^{3}} =  \frac{10 g}{volume}

Volume of acetone = 12.72 cm^{3}

Thus, volume of acetone is more than volume of water because the density of acetone is lower.

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The major use of carbon dioxide​
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Carbon dioxide is used as a refrigerant, in fire extinguishers, for inflating life rafts and life jackets, blasting coal, foaming rubber and plastics, promoting the growth of plants in greenhouses, immobilizing animals before slaughter, and in carbonated beverages.

3 0
2 years ago
Determine whether a precipitate will form when 0.96g of Na2CO3 is combined with 0.20g BaBr2 in a 10 L solution.
Mashcka [7]

Answer:

Qsp > Ksp, BaCO3 will precipitate

Explanation:

The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M

Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles

concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M

Hence;

[Ba^2+] = 6.7 * 10^-5 M

[CO3^2-] = 9.1 * 10^-4 M

Qsp = [6.7 * 10^-5] [9.1 * 10^-4]

Qsp = 6.1 * 10^-8

But, Ksp for BaCO3 is 5.1*10^-9.

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3 years ago
Which combination of factors is most suitable for increasing the electrical conductivity of metals
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What is the vapor pressure of a liquid at 305.03 K if its ∆Hvap = 28.9 kJ/mol and its normal boiling point is 341.88 K?
ASHA 777 [7]

<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm

<u>Explanation:</u>

To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:

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

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm

P_2 = pressure of the liquid = ?

\Delta H_{vap} = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 341.88 K

T_2 = final temperature = 305.03 K

Putting values in above equation, we get:

\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm

Hence, the vapor pressure of the liquid is 0.293 atm

5 0
2 years ago
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