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3 years ago

What is the density of krypton gas (MM = 83.8 g/mol) at 3.00 atm and 100. °C? g/L

Chemistry

1 answer:

ycow [4]3 years ago

4 0

Answer:

8.22 g/dm³

Explanation:

Applying,

PV = nRT............... Equation 1

Where P = pressure of krypton, V = Volume of krypton, n = number of moles, R = molar gas constant, T = Temperature

But,

number of mole (n) = mass(m)/molar mass(m')

n = m/m'.............. Equation 2

Substitute equation 2 into equation 1

PV = mRT/m'

P = (m/V)(RT)/m'................ Equation 3

Also,

Density (D) = Mass(m)/Volume(V)

D = m/V.............. Equation 4

Substitute equation 4 into equation 3

P = DRT/m'

D = Pm'/RT.................... Equation 5

From the question,

Given: P = 3.00 atm, T = 100°C = (273+100) = 373 K, m' = 83.8 g/mol

Constant: R = 0.082atm.dm³.K⁻¹.mol⁻¹

Substitute these values into equation 5

D = (3×83.8)/(0.082×373)

D = 251.4/30.586

D = 8.22 g/dm³

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3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would

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Answer:

Percent yield = 89.1%

Explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

<em>Moles KI -Molar mass: 166.0028g/mol-</em>

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

<h3>Percent yield = 89.1%</h3>

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3 years ago

How many lead atoms are in a cube of lead that has a volume of 2.18 cm3 if the density of lead is 11.3 g/cm3?

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3 years ago

What is 20 times 200000000

iren [92.7K]

Answer: 4000000000

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3 years ago

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One is rows the other is columns

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3 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

Katarina [22]

Answer:

empirical formula: $H_2SO_4$

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: $H_2SO_4$ .

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3 years ago