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3 years ago

What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

6 0

Answer:

2.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 0.109 mole

Pressure (P) = 0.98 atm

Temperature (T) = 307 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

The volume of the helium gas can be obtained by using the ideal gas equation as follow:

PV = nRT

0.98 × V = 0.109 × 0.0821 × 307

0.98 × V = 2.7473123

Divide both side by 0.98

V = 2.7473123 / 0.98

V = 2.8 L

Thus, the volume of the helium gas is 2.8 L.

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Naphthalene is soluble in diethyl ether, but it is insoluble in water regardless of the solution pH. Explain why this compound c

Mademuasel [1]

Answer:

Naphthalene is a non polar substance

Explanation:

In chemistry, the principle of like dissolves like is the over aching principle that controls the dissolution of one substance in another.

Naphthalene dissolves in diethyl ether because diethyl ether is a non polar solvent just as naphthalene is a nonpolar substance.

Since water is a polar solvent, it can not dissolves naphthalene at any pH because naphthalene is a nonpolar substance.

5 0

4 years ago

Please help me.

kirza4 [7]

Taking into account the definition of density, the density of the sample is 0.5 $\frac{g}{mL}$ .

It is necessary yo know that density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density allows you to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

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

In this case, you know:

mass= 6 g
volume= 12 mL

Then, replacing in the definition of density:

$density=\frac{6 g}{12 mL}$

Solving:

density= 0.5 $\frac{g}{mL}$

Finally, the density of the sample is 0.5 $\frac{g}{mL}$ .

Learn more about density:

7 0

3 years ago

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$