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3 years ago

4.Calculate the volume of 40g of Helium (He) at rtp.

Chemistry

1 answer:

Oksana_A [137]3 years ago

4 0

Answer:

2.24dm³

Explanation:

Given parameters:

Mass of He = 40g

Unknown:

Volume of Helium = ?

Solution:

To solve this problem, we convert the given mass to number of moles.

Number of moles = $\frac{mass}{molar mass}$

molar mass of He = 4g/mol

Number of moles = $\frac{4}{40}$ = 0.1mole

So;

1 mole of gas at rtp occupies a volume of 22.4dm³

0.1 mole of He will occupy a volume of 0.1 x 22.4 = 2.24dm³

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If the distance between two objects of the same mass decreased the force of gravity between them would

Anni [7]

Answer:

The force of gravity between two objects will decrease as the distance between them increases.

Explanation:

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3 years ago

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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

$0.400 = 9 k$

=> $k = 0.044\ L/ mol \cdot hr^{-1}$

For $C_Z = 0.42M$

$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

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3 years ago

1. A given mass of air has a volume of 6.00 L at 80.0°C. At constant pressure, the temperature is

earnstyle [38]

3 L will be the final volume for the gas as per Charle's law.

Answer:

Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

V∝ T

Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

$\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }$

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.

$\frac{6}{80}=\frac{V_{2} }{40}$

$V_{2}=\frac{6*40}{80}=3 L$

So, 3 L will be the final volume for the gas as per Charle's law.

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3 years ago

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olga nikolaevna [1]

Answer:

Heat or electricity in the atmasphere

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2 years ago

Ethyne gas combusts with oxygen gas according to the following reaction: Calculate the volume, in mL of CO2 produced when 73 g o

Bezzdna [24]

Answer: Volume of CO2 is 89127 mL

Explanation: The reaction that takes place is: C2H2 + O2 --> CO2 + H2O

The amount of C2H2 that react allow us to predict the amount of CO2 that will be obtained

$mol CO2 = 73gC2H2 .\frac{1 mol C2H2}{26gC2H2} . \frac{2mol CO2}{4mol C2H2} = 5,6 mol CO2$

26g/1mol is molar mass of C2H2 and 2/4 is the molar relation between CO2 and C2H2 in this reaction. Canceling units, at the end mol of CO2 are obtained

Now with the moles of CO2 and the ideal gases equation is possible to calculate the volumen occupied by the gas.

PV = RnT where P: pressure, V: volume, R: ideal gas constant, n: moles and T: temperature expressed in K (add 273,15 to °C temperature: 37,4°C + 273,15 = 310,55K)

V= RnT/P

$V= \frac{0,08206 atmL/molK . 5,6 mol. 310,55 K}{1,6 atm} =89,127 L$

To express volume in mL multiply the L result by 1000 which equals 89127 mL

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3 years ago