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If 0.896 g of a gas occupies a 250 mL flask at 20°C and 760 mm Hg of pressure, what is the molar mass of the gas?

salantis [7]

Answer:

86.2 g/mol

Explanation:

Before you can find the molar mass, you first need to calculate the number of moles of the gas. To find this value, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

After you convert the volume from mL to L and the temperature from Celsius to Kelvin, you can use the equation to find the moles.

P = 760 mmHg R = 62.36 L*mmHg/mol*K

V = 250 mL / 1,000 = 0.250 L T = 20 °C + 273.15 = 293.15 K

n = ? moles

PV = nRT

(760 mmHg)(0.250 L) = n(62.36 L*mmHg/mol*K)(293.15 K)

190 = n(18280.834)

0.0104 = n

The molar mass represents the mass (g) of the gas per every 1 mole. Since you have been given a mass and mole value, you can set up a proportion to determine the molar mass.

$\frac{?grams}{1 mole} =\frac{0.896grams}{0.0104moles}$ <----- Proportion

$?grams(0.0104moles) = 0.896$ <----- Cross-multiply

$?grams = 86.2$ <----- Divide both sides by 0.0104

8 0

2 years ago

Soluble or Insoluble in water: Dirt

Varvara68 [4.7K]

Answer:

Insoluble

Explanation:

Like other non-polar molecules such as petrol, wax and grease, most food and dirt is not soluble in water.

https://www.primaryconnections.org.au/sites/all/modules/primaryconnections/includes/SBR/data/Chem/sub/soap/soap.htm#:~:targetText=Like%20other%20non%2Dpolar%20molecules,is%20not%20soluble%20in%20water.

5 0

3 years ago

A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or

Pavlova-9 [17]

Answer:

(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.

(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.

(c) The expected number of fatalities on a average chemical plant are one in 12500 years.

Explanation:

Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.

In 300 years we have 600,000 hours.

With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.

If we have 2000 work hours a year, it is expected 0.0033 deaths/year.

$1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year$

The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).

In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:

$FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8} hours=1.67*10^{2} =167$

A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.

If we assume FAR=4, as in an average chemical plant, we expect

$4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}$

This is equivalent to say

$\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}$

The expected number of fatalities on a average chemical plant are one in 12500 years.

4 0

3 years ago

The diagram shows a balloon that was rubbed with a piece of cloth what is the change

Ede4ka [16]

Answer: the electric charge

Explanation:

4 0

3 years ago

What is the interval used on the y-axis of this graph?

bezimeni [28]

Answer:

5

Explanation:

the numbers go up by 5

7 0

3 years ago

ASAP please and Thankyou​​

ASAP please and Thankyou