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3 years ago

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A flexible container at an initial volume of 3.10 L contains 9.51 mol of gas. More gas is then added to the container until it r

eaches a final volume of 19.1 L . Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

1 answer:

Schach [20]3 years ago

7 0

Answer:

49.09 moles of gas are added to the container

Explanation:

Step 1: Data given

Initial volume = 3.10 L

Number of moles gas = 9.51 moles

The final volume = 19.1 L

The pressure, temperature remain constant

Step 2: Calculate number of moles gas

V1/n1 = V2/n2

⇒with V1 = the initial volume of the gas = 3.10 L

⇒with n1 = the initial number of moles = 9.51 moles

⇒with V2 = the increased volume = 19.1 L

⇒with n2 = the final number of moles gas

3.10L / 9.51 moles = 19.1 L / n2

n2 = 58.6 moles

The new number of moles is 58.6

Step 3: calculate the number of moles gas added

Δn = 58.6 - 9.51 = 49.09 moles

49.09 moles of gas are added to the container

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Calculate the number of molecules in 46.0 grams of water

Natalka [10]

Answer:

Explanation:

Since water has a chemical formula of H2O , there will be 2 moles of hydrogen in every mole of water. In one mole of water, there will exist approximately 6.02⋅1023 water molecules.

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3 years ago

Find the ph of a buffer that consists of 0.18 m ch3nh2 and 0.73 m ch3nh3cl (pkb of ch3nh2 = 3.35)?

ozzi

Hello!

First, we need to determine the pKa of the base. It can be found applying the following equation:

$pKa=14-pKb=14-3,35=10,65$

Now, we can apply the Henderson-Hasselbach's equation in the following way:

$pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04$

So, the pH of this buffer solution is 10,04

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8 0

3 years ago

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, in grams, must react to produce 50.

snow_lady [41]

112.5 g. The production of 50.00 g O2 requires 112.5 g H2O.

a) Write the partially balanced equation for the decomposition of water.

MM = 18.02 32.00

2H2O → O2 + …

Mass/g = 50.00

b) Calculate the <em>moles of O2
</em>

Moles of O2 = 50.00 g O2 × (1 mol O2/16.00 g O2) = 3.1250 mol O2

c) Calculate the <em>moles of water</em>

Moles of H2O = 3.1250 mol O2 × (2 mol H2O/1 mol O2)

= 6.2500 mol H2O

d) Calculate the mass of water

Mass of H2O = 6.2500 mol H2O × (18.02 g H2O/1 mol H2O)

= 112.5 g H2O

5 0

3 years ago

Combustion analysis of a 13.42-g sample of estriol (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 1

meriva

Answer:

C18H24O3

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of estriol = 13.42g

Mass of CO2 = 36.86g

Mass of H2O = 10.06g

Molar mass of estriol = 288.38g/mol

Step 2:

Determination of the mass of Carbon (C), Hydrogen (H) and Oxygen (O) present in the compound. This is illustrated below:

For Carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 36.86 = 10.05g

For Hydrogen, H:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 10.06 = 1.12g

For Oxygen, O:

Mass of O = 13.42 – (10.05 + 1.12) = 2.25g

Step 3:

Determination of the empirical formula for estriol. This is illustrated below:

C = 10.05g

H = 1.12g

O = 2.25g

Divide by their molar mass

C = 10.05/12 = 0.8375

H = 1.12/1 = 1.12

O = 2.25/16 = 0.1406

Divide by the smallest i.e 0.1406

C = 0.8375/0.1406 = 6

H = 1.12/0.1406 = 8

O = 0.1406/0.1406 = 1

Therefore, the empirical formula for estriol is C6H8O

Step 4:

Determination of the molecular formula for estriol. This is illustrated below:

Molecular formula is simply a multiple of the empirical formula i.e

Molecular formula => [C6H8O]n

[C6H8O]n = 288.38g/mol

[(12x6) + (8x1) + 16]n = 288.38

[72 + 8 + 16]n = 288.38

96n = 288.38

Divide both side by 96

n = 288.38/96 = 3

Molecular formula => [C6H8O]n

=> [C6H8O]n

=> [C6H8O]3

=> C18H24O3

Therefore, the molecular formula for estriol is C18H24O3

4 0

3 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago