All categories

3 years ago

Gases with high molecular weights diffuse more slowly than gases with lower molecular weights.

2 answers:

7 0

The answer to this item is TRUE. This can be explained through the Graham's law. This law states that the rate at which gases diffuse is inversely proportional to the square root of their densities which is also related to their molecular masses.

serg [7]3 years ago

6 0

Answer:

The given statement is true.

Explanation:

Graham's Law:

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molecular mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

Higher the molecular mass lower will be effusion rate.
Lower the molecular mass higher will be the effusion rate.

You might be interested in

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

The half-life of C-14 is 5470 years. If a particular archaeological sample has one-quarter of its original radioactivity remaini

-Dominant- [34]

10940? Sorry if it isn’t correct

3 0

3 years ago

Read 2 more answers

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago

A. Draw a graph that shows the relative concentration of BTB from the center of the

rosijanka [135]

Answer:

did you get the answer???

Explanation:

I really need help

6 0

3 years ago

3. Oblicz stężenie procentowe 423 g roztworu zawierającego 17 g rozpuszczonego siarczanu (VI) miedzi (II). Zapisz obliczenia i o

goldfiish [28.3K]

Answer:

17 / 423 = 0.04018912529

~ 4%

proszę bardzo

6 0

4 years ago