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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.

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How many moles are there in 13g of KMnO4?

Hoochie [10]

Answer:

0.08 moles

Explanation:

Relative atomic mass of KMnO4 = K + Mn + O4 = 39 + 55 + 16 x 4 = 158

=> Moles in 13g of KMnO4 = $\frac{Mass}{Relative-atomic-mass}$ = $\frac{13}{158}$ = 0.082278.. = 0.08 moles

3 0

3 years ago

Read 2 more answers

Mg + 2AgNO3 -> 2Ag + Mg(NO3)2

9966 [12]

Answer:

Explanation:

We'll assume there is an excess of silver nitrate, so that all 12.0 moles of the magnesium (Mg) will react.

The balanced equation tells us we'll obtain 2 moles of Ag for every 1 mole of magnesium, for a molar ratio of 2/1.

Starting with 12.00 moles Mg, we would therefore hope to find twice that, or 24.00 moles of Ag.

To convert to grams, find the molar mass of Ag from the periodic table.

Ag has a molar mass of 107.9 (to 4 sig figs) grams/mole.

(24.00 moles)*(107.9 grams/mole) = 2590 grams (4 sig figs)

Hands off, it's mine.

5 0

2 years ago

What is the volume of 40g of sugar given it’s listed density

lora16 [44]

Answer:

25.157 cm³

Explanation:

Data Given:

Mass of Sugar (m) = 40g

Density of sugar given in literature = 1.59 g/cm³

Volume of Sugar = ?

The formula will be used is

d = m/v ........................................... (1)

where

D is density

m is the mass

v is the volume

Rearrange the Equation (1)

d x v = m

v = m/ d ................................................ (2)

put the given values in Equation (2)

v = 40g / 1.59 g/cm³

v = 25.157 cm³

volume of 40 g of sugar = 25.157 cm³

8 0

3 years ago

A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 m

DerKrebs [107]

The pH of the solution after the addition of 100. 0 ml of HNO3. The kb of NH3 is 1. 8 × 10-5. So, pH is 10.9 basic .

This neutralization occurs as the acid is added to the base:

NH3(aq)+ HNO3(aq)= NH4NO3(aq)+ H2O(l)

The initial moles of NH3present is given by,

nNH3= c×v= 0.10× 100/1000= 0.01m

The number of moles of

HNO3 added is given by:

nHNO3= c×v= 0.10× 100/1000= 0.001

It is clear from the equation that the acid and base react in a 1:1 molar ratio. So, the no. moles of NH3remaining will be 0.01 - 0.001= 0.009

The total volume is now

100.0+100.0= 200.0x cm3

The concentration of NH3 is given by,

[NH3]= c/v= 0.001/200/1000= 0.05x mol/l .

pOH= 12 (pKb− logb)

where b is the base's concentration.

Given that there is little dissociation, we can roughly compare this to the starting concentration.

pKb= −logKb= −log(1.8×10−5)= 4.744

pOH= 3.056

At 25∘xC, we know that, pH+ pOH=14

pH= 14− 3.056= 10.9

To know more about Equilbrium, visit-brainly.com/question/14366127

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8 0

1 year ago

This is my chemistry worksheet. It's a new topic my teacher will barely review with us. Help please???

TEA [102]

So to balance an equation, you need to get the same amount of each type of element on either side of the --> . So you pretty much are given the subscripts in the equations and you need to add coefficients (just normal numbers) in front of any formula that needs it, keeping anything balance.
$KCl_{3} + O_{2} -\ \textgreater \ KCl_{3}$
turns into
$2KCl_{3}+ 3O_{2} -> 2KCl_{3}$

These coefficient numbers are the molar ratios, so 2 moles of KCl3 for every 3 moles of O2 so 1. 3:2

Then you can use these ratios of find out how many moles of one thing are needed if you are given the amount of another.
$\frac{moles of element 1}{cofficient 1} = \frac{moles of element 2}{cofficient 2}$
and use cross multiplication to solve for whatever you don't know

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6 0

3 years ago