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3 years ago

How many atoms are in a 3.5 mole sample of the element lithium?

Chemistry

1 answer:

seraphim [82]3 years ago

4 0

Answer:

2.1 x 10^24

Explanation:

The number of atoms in a mole will always be the same. The fact that it's lithium is completely irrelevant.

One mole of anything is 6.022 x 10^23 atoms

So if we need to find the number of atoms in 3.5 mol we just need to multiply Avogadro's number by 3.5

3.5 x 6.022 x 10^23

= 21.077 x 10^23

To make it proper scientific notation the first number can't be greater than 10 so let's move the decimal place and increase the exponent

2.1077 x 10^24

Remove the numbers that aren't significant figures

2.1 x 10^24

Boom there's your answer

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4 years ago

A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is

Airida [17]

Answer:

the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C

Explanation:

Given the data in the question;

First we calculate the Volume of the steel cylinder;

V = πr²h

radius r = Diameter / 2 = 27 cm / 2 = 13.5 cm

height h = 32.4 cm

so we substitute

V = π × ( 13.5 cm )² × 32.4 cm

V = π × 182.25 cm × 32.4 cm

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V = 18.551 L

given that; maximum safe pressure P = 3.10 MPa = 30.5946 atm

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molar mass of carbon monoxide gas is 28.010 g/mol

moles of carbon monoxide gas n = 218 gram / 28.010 g/mol = 7.7829 mol

we know that;

PV = nRT

solve for T

T = PV / nR

we know that gas constant R = 0.0820574 L•atm•mol⁻¹ K⁻¹

so we substitute

T = ( 30.5946 × 18.551 ) / ( 7.7829 × 0.082 )

T = 567.5604 / 0.6381978

T = 889.317387 K

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Therefore, the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C

6 0

3 years ago

A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release

Marrrta [24]

Answer:

$V_2=12.1L$

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

$T_1=20+273=293K\\\\T_2=-40+273=233K$

Then, we obtain:

$V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L$

Best regards!

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