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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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Answer:

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3 years ago

3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O

liubo4ka [24]

24.25 moles of NO can be produced using 97 moles of HNO3.

<h3>What is balanced chemical equation?</h3>

Equal numbers of atoms from various elements are present in both the reactants and the products in balanced chemical equations. Varied elements' atom counts in the reactants and products of unbalanced chemical equations are different.

3 Cu + 8HNO3 g → 3 Cu(NO3)2 + 2 NO + 4 H2O

The number of moles consumed can be calculated using comparing with coefficients in the balanced reaction .

So , from above eq we get that 8 moles of HNO3 are consumed to make 2 moles of NO.

⇒ 8 HNO3⇔2 NO

⇒ 1 HNO3⇔ 1/4 NO

This means that for each mole of HNO3 produces 1/4 moles of NO.

So , for 97 moles of HNO3 , $\frac{1}{4} *97$ moles of NO can be made,

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Lean more about balanced reactions here brainly.com/question/26694427

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2 years ago

Read 2 more answers

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

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3 years ago

Which element is identified by the chemical symbol C?

GrogVix [38]

Answer:

B Carbon

Explanation:

I really hopes this helps you!

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3 years ago