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Digiron [165]
3 years ago
9

Question 4 A mixture of xenon and oxygen gas is compressed from a volume of to a volume of , while the pressure is held constant

at . Calculate the work done on the gas mixture. Round your answer to significant digits, and be sure it has the correct sign (positive or negative).
Chemistry
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

W = - 1218 atm L

Explanation:

Complete question

A mixture of krypton and oxygen gas is expanded from a volume of 23.0L to a volume of 81.0L, while the pressure is held constant at 21.0atm. Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.

Solution

As we know that -

Work done is equal to the product of pressure applied and change in the difference of volume of two liquids.

This can be represented mathematically as follows -

W = -P*(V_2-V_1)\\

Where W represents the work done

P represents the change in pressure

V_2 is the final volume  and

V_1 is the initial volume

Substituting the given values in above equation, we get -

W = - 21 * (81 -23)\\

W = - 1218 atm L

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Catalyst
Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

\tt \dfrac{4}{5}\times 0.5=0.4

volume NO at 1273 K and 1 atm

\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L

b. 15 L NH3 at STP ( 1mol = 22.4 L)

\tt \dfrac{15}{22.4}=0.67~mol

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

\tt \dfrac{6}{4}\times 0.67=1

mass H2O(MW = 18 g/mol) :

\tt mass=mol\times MW=1\times 18=18~g

c. mol NO at 1273 K and 1 atm :

\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34

mol ratio of NO : O2 = 4 : 5, so mol O2 :

\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

\tt 0.425\times 22.4=9.52~L

5 0
3 years ago
A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the
yarga [219]

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X  1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of  H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

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2 years ago
How was Bohr’s atomic model different from Rutherford’s atomic model?
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C is the answer.

It says on the third picture that Bohr refined Rutherford's model by giving distinct orbits for the electrons with distinct radii.


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