Ask question

Ask question

All categories

2 years ago

12

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

f 102 mL/s. Calculate the molecular mass of the unknown gas.

1 answer:

Ostrovityanka [42]2 years ago

4 0

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

You might be interested in

A sample of nitrogen gas has a mass of 48.6 grams. How many N2 molecules are there in the sample? molecules Submit Answer &

marysya [2.9K]

Answer:

There are 1.05 x 10²⁴ molecules in 48.6 g N₂

Explanation:

1 mol of N₂ has a mass of (14 g * 2) 28 g.

Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.

Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be

(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05 x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).

6 0

2 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

2 years ago

Draw the alcohol that is the product of the reduction of 3-methylpentanal.

natali 33 [55]

Answer: The product from the reduction reaction is

CH3-CH2-CH(CH3)-CH2-CH2OH

IUPAC name; 3- Methylpentan-1-ol

Explanation:

Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.

Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal

CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;

3-methylpenta-1-ol .

The structure of the product is:

CH3-CH2-CH(CH3)-CH2-CH2OH

7 0

2 years ago

In a voltaic cell, the salt bridge acts to:

AleksAgata [21]

<span>Provide a direct contact between the oxidation and reduction electrodes - A</span>

6 0

2 years ago

Read 2 more answers

A sample of ethanol (C2H6O) has a mass of 0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorime

Keith_Richards [23]

Answer: $4.994\times 10^{-3}$ moles

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass , occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

Given mass of ethanol = 0.2301

Molar mass of ethanol = 46.07 g/mol

$\text{Number of moles of ethanol}=\frac{0.2301g}{46.07g/mol}=4.994\times 10^{-3}$

Thus there are $4.994\times 10^{-3}$ moles of ethanol are present in the sample.

7 0

2 years ago

Read 2 more answers