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1 year ago

Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.

1 answer:

6 0

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation is given below.

$PV = nRT$

The equation can be written as

$\rm PV = \dfrac{m}{M} \ RT\\\\\\MPV = mRT$

Where M is the molar mass, P is the pressure, V is the volume, R is the universal gas constant, T is the temperature, and m is the mass of the gas.

Then the molar mass of a 0.622-gram sample of gas has a volume of 2.4 L at 287 K and 0.850 atm.

V = 2.4 L = 0.0024

P = 0.85 atm = 86126.25 Pa

T = 287 K

m = 0.622

R = 8.314

Then we have

$\begin{aligned} \rm M \times 86126.25 \times 0.0024&= 0.622 \times 8.314 \times 287\\\\\rm M &= 7.18 \end{aligned}$

Then the molar mass of the gas is 7.18.

More about the ideal gas equation link is given below.

brainly.com/question/4147359

#SPJ1

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Please help , thanks

Bogdan [553]

Answer:

u want to find volume?

if so then, it is = 150cm³

Step-by-step explanation:

1/3×5×9×10=150cm³

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2 years ago

A sample of blood pressure measurements is taken for a group of adults, and those values (mm Hg) are listed below. The values

nika2105 [10]

Answer:

For systolic pressure data:

$\bar X =127.2\\\\S=17.91668\\\\CV=\frac{17.91668}{127.2}=0.14085$

For diastolic pressure data:

$\bar X =73.5\\\\S=12.54\\\\CV=\frac{73.5}{12.54}=0.17061$

Systolic pressure is slightly less variable, among individuals in the sample, than diastolic pressure.

Step-by-step explanation:

The coefficient of variation is defined as the percentage relative variation of a set of data with respect to its average. And it is calculated like this:

$CV=\frac{\bar X}{S}$

$S =\sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})^2$

$\bar X={\frac{1}{n} \sum_{i=1}^n x_i$

For systolic pressure data:

$\bar X =127.2\\\\S=17.91668\\\\CV=\frac{17.91668}{127.2}=0.14085$

For diastolic pressure data:

$\bar X =73.5\\\\S=12.54\\\\CV=\frac{12.54}{73.5}=0.17061$

It is observed that the systolic pressure shows greater standard deviation but less coefficient of variation. This is due to the greater magnitude of its measurement scale.

Systolic pressure is slightly less variable, among individuals in the sample, than diastolic pressure.

7 0

2 years ago

Can someone please help with #16

ollegr [7]

Answer:

∠ WZX = 50°

XW is not an altitude.

Step-by-step explanation:

16. See the attached figure.

XW is the angle bisector of ∠ YXZ, hence, ∠ WXY = ∠ WXZ

Now, given that ∠ YXZ = 8x + 34 and ∠ WXY = 10x - 13

Hence, ∠ YXZ = 2 ∠ WXY

⇒ 8x + 34 = 2(10x - 13)

⇒ 8x + 34 = 20x - 26

⇒ 12x = 60

⇒ x = 5.

Hence, ∠ XZY = ∠ WZX = 10x = 50° (Answer)

Now, ∠ WXZ = ∠ WXY = 10x - 13 = 37°

Hence, from Δ WXZ,

∠ WZX + ∠ WXZ + ∠ XWZ = 180°

⇒ 50° + 37° + ∠ XWZ = 180°

⇒ ∠ XWZ = 93° ≠ 90°

Hence, XW is not an altitude. (Answer)

6 0

2 years ago

Janet can make 3/5

IceJOKER [234]

I need help with this problem too

7 0

2 years ago

Why does a logarithmic equation sometimes have an extraneous solution?

jonny [76]

The main reason behind this is using properties of logarithm .

when solving

log_2(x+2)+log_2(x-3)=3

There you use multiplication property and make addition to multiplication then you get extraneous solution because in plenty of cases they occur

(-2)²=(2)²

But

-2≠2

Same happens in case of logarithm

6 0

1 year ago

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