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kupik [55]
3 years ago
13

In which of the substances in Figure 3-1 are the forces of attraction among the particles so weak that they can be ignored under

ordinary conditions?
A. Substance A
B. Substance B
C. Substance C
D. Substance A, B, and C

Chemistry
2 answers:
lozanna [386]3 years ago
5 0

Answer: Option (C) is the correct answer.

Explanation:

Forces of attraction between the particles will be weak when they are far apart from each other. Hence, only then the molecules will be able to move rapidly and they can be ignored at ordinary conditions.

For example, in gases molecules are held together by weak Vander waal forces.

Whereas solids and liquids have more force of attraction between their molecules as compared to gases as they cannot be ignored under ordinary conditions.

Thus, we can conclude that in substance C forces of attraction among the particles so weak that they can be ignored under ordinary conditions.

yawa3891 [41]3 years ago
3 0
Substance C can be compared with gaseous particles in which intermolecular forces are so weak because particles are far from each other. 
Hence option C is correct.
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a sample of helium occupies a volume of 101.2 mL at a pressure of 790 mmHg. at what pressure would the volume be 120 mL?
AveGali [126]

Answer : The final pressure will be, 666.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 790 mmHg

P_2 = final pressure = ?

V_1 = initial volume = 101.2 mL

V_2 = final volume = 120 mL

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

790mmHg\times 101.2mL=P_2\times 120mL

P_2=666.2mmHg

Therefore, the final pressure will be, 666.2 mmHg

6 0
3 years ago
(a) In a gaseous sample of oxygen atoms and diatomic oxygen in equilibrium at 298K andunder a pressure of 1atm, what fraction of
ZanzabumX [31]

Answer:

Explanation:see below as shown from pics

7 0
3 years ago
An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is
aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x (\frac{1 L}{1000 mL}) = .236 L

Pressure= 740 mm Hg x (\frac{1 atm}{760 mm Hg})= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= \frac{mass}{moles} or MM= \frac{m}{n}. They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n=\frac{PV}{RT}, where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n=\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}

n= 0.009 mol

Let's look back at MM= \frac{m}{n} and plug in what we know.

MM= \frac{0.443 g}{0.009 mol}

MM= 49.2 g/mol

3 0
3 years ago
C. How many moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg?
olya-2409 [2.1K]

Answer: There are 4.45\times 10^{-4}moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 895 mm Hg= 1.18 atm  (760 mm Hg= 1 atm)

V = Volume of gas = 9.55 ml = 0.00955 L   (1 L=1000ml)

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =35^0C=(35+273)K=308K

n=\frac{PV}{RT}

n=\frac{1.18atm\times 0.00955L}{0.0821L atm/K mol\times 308K}=4.46\times 10^{-4}moles

Thus there are 4.45\times 10^{-4}moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

5 0
3 years ago
If you were to test how much time it takes to melt a piece of ice in warm water, what are
77julia77 [94]
Time, Ice size, water temp
8 0
3 years ago
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