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3 years ago

Calculate the molar mass of a gas at 506 torr and 99°c if 206 ng of gas occupies 0.206 ul. (

Chemistry

1 answer:

VikaD [51]3 years ago

7 0

PV = nRT

and n = mass / mw

PV = (mass / mw) RT

mw = mass RT / PV

P = 352 torr = 352 torr x (1 atm / 760 torr) = 0.463 atm

V = 0.206x10^-6 L

R = 0.0821 L atm /moleK

T = 45C = 45+273 = 318 K

mass = 243x10^-9 g

mw = (243x10^-9 g)x(0.0821 L atm /moleK)x(318 K) / [(0.463 atm)x(0.206x10^-6 L)]

mw = 66.5 g/mole

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Alina [70]

The empirical formula for this vitamin : C₃H₄O₃

<h3>Further explanation </h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

Determine the mass ratio of the constituent elements of the compound.
Determine the mole ratio by dividing the percentage by the atomic mass

Mass of C in CO₂ :(MW C = 12 g/mol, CO₂=44 g/mol)

$\tt \dfrac{12}{44}\times 1.4991=0.409~g$

Mass of H in H₂O :(MW H = 1 g/mol, H₂O = 18 g/mol)

$\tt \dfrac{2.1}{18}\times 0.4092=0.0455~g$

Mass O = Mass sample - (mass C + mass H) :

$\tt 1-(0.409+0.0455)=0.5455~g$

mol ratio C : H : O =

$\tt \dfrac{0.409}{12}\div \dfrac{0.0455}{1}\div \dfrac{0.5455}{16}\\\\0.0341\div 0.0455\div 0.0341\rightarrow 1\div 1.33\div 1=3\div 4\div 3$

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3 years ago

.an open flask contains 0.200 mol of air. atmospheric pressure is 745 mmhg and room temperature is 68˚f. how many moles are pres

aleksklad [387]

0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.

What is ideal gas constant ?

The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
The combined gas law relates pressure, volume, and temperature of a gas.

We simple use this formula-

The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n = of moles (mol) R = the Ideal Gas Law Constant.

68F = 298.15K

V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters

n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)

n = 0.219 moles

Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.

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2 years ago

If heat transfers from the system (solute) to the surroundings (solvent), then AH is

balu736 [363]

Answer:

B) exothermic.

Explanation:

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In this case, we need to keep in mind that exothermic reactions release heat, so they increase the temperature as the final energy is less than the initial energy; in contrast, endothermic reactions absorb heat, so they decrease the temperature as the final energy is greater than the initial energy.

In such a way, when a dissolution process shows off a negative enthalpy of dissolution, we infer it is an exothermic process due to the aforementioned; therefore, the answer is:

B) exothermic .

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3 years ago

A student sees an absorbance a=1.140 for his solution that has a concentration of c=1.50*10-4 m using 0.50 cm cuvette. what is t

777dan777 [17]

The molar extinction coefficient is 15,200 $M^{-1} cm^{-1}$ .

The formula to be used to calculate molar extinction coefficient is -

A = ξcl, where A represents absorption, ξ refers molar extinction coefficient, c refers to concentration and l represents length.

The given values are in required units, hence, there is no need to convert them. Directly keeping the values in formula to find the value of molar extinction coefficient.

Rewriting the formula as per molar extinction coefficient -

ξ = $\frac{A}{cl}$

ξ = $\frac{1.140}{1.5*10^{-4}*0.5 }$

Performing multiplication in denominator to find the value of molar extinction coefficient

ξ = $\frac{1.140}{0.000075}$

Performing division to find the value of molar extinction coefficient

ξ = 15,200 $M^{-1} cm^{-1}$

Hence, the molar extinction coefficient is 15,200 $M^{-1} cm^{-1}$ .

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1 year ago

Can some atoms exceed the limits of the octet rule in bonding? If so, give an example.

harkovskaia [24]

Answer:

Yes. Example: Sulfur hexafluoride (SF₆) molecule

Explanation:

According to the octet rule, elements tend to form chemical bonds in order to have 8 electrons in their valence shell and gain the stable s²p⁶ electronic configuration.

However, this rule is generally followed by main group elements only.

Exception: SF₆ molecule

In this molecule, six fluorine atoms are attached to the central sulfur atom by single covalent bonds.

Each fluorine atom has 8 electrons in their valence shells. Thus, it follows the octet rule.

Whereas, there are 12 electrons around the central sulfur atom in the SF₆ molecule. Therefore, sulfur does not follow the octet rule.

Therefore, the SF₆ molecule is known as a hypervalent molecule or expanded-valence molecule.

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3 years ago