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3 years ago

Which statement best explains why different gases effuse at different rates?

Chemistry

2 answers:

arlik [135]3 years ago

6 0

Your answer is A. <span>At a given pressure, gas particles with smaller masses move faster.

Hope this helps.

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matrenka [14]3 years ago

6 0

Answer: Different gases effuse at different rates because at a given temperature, gas particles with smaller masses move faster.

Explanation: The motion of particles in a gas depends on the kinetic energy of the gas particles which further depend on the temperature of the gas. Thus, the rate of gas given by Graham's law can only be stated when the two gases are at fixed temperature.

Graham's Law: This law states that the rate of effusion of gas is inversely proportional to the square root of its mass.

Mathematically,

$\frac{(Rate)_1}{(Rate)_2}=\sqrt{\frac{M_2}{M_1}}$

From the above relation, the gas particle with smaller mass will effuse faster at a given temperature.

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When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

4 years ago

What kind of intermolecular forces act between two ammonia molecules

iragen [17]

Answer:

extensive hydrogen bonding

Explanation:

The high boiling points of water, hydrogen fluoride (HF) and ammonia (NH3) is an effect of the extensive hydrogen bonding between the molecules. The London dispersion force is caused by random and temporary changes in the polarity of atoms, caused by the location of the electrons in the atoms' orbitals.

Hope this helps :)

6 0

2 years ago

A sample of glucose contains 1.250x10^21 carbon atoms, how many atoms of hydrogen does it contain?

dybincka [34]

Answer:

Hydrogen = 2.5 * 10^21

Explanation:

Chemical Formula Glucose: C₆H₁₂O₆

One of the ways you could do this is to notice that for every carbon atom there are two Hydrogen atoms. You can state this more formally by using the formula to set up a ratio: 12/6 = hydrogen to Carbon

So if there are 1.250 * 10^21 Carbon atoms in the Glucose sample, then there will be twice as many hydrogen atoms.

H = 2 * 1.25 * 10^21 = 2.5 * 10^21 atoms

You could do this more formally by setting up a proportion.

6 Carbon / 12 Hydrogen = 1.25*10^21 / x Cross Multiply

6*x = 12 * 1.25*10^21 Combine the right

6x = 1.5 * 10^22 Divide by 6

x = 2.5 * 10^21

5 0

2 years ago

I need help ASAP

vova2212 [387]

Answer:

1.4 g/cm3

Explanation:

Density = Mass/Volume

Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

8 0

3 years ago

Explain why ionic compound do not form molecules.

Darina [25.2K]

Answer:

Because a molecule, by definition, has a valence of zero

(neutral charge, stable). Also by definition, an ion has a positive

or negative charge or valence and is not stable.

Explanation:

7 0

3 years ago