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jeka94
3 years ago
6

The volume occupied by 0.25 mol of sulfur dioxide at rtp ​

Chemistry
1 answer:
Vladimir79 [104]3 years ago
8 0

Answer:

V = 0.25 x 22.4 Liters

Explanation:

1 mole of any gas occupies 22.4 Liters at STP.

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An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi
Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr  are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM →  54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3  . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3  .  11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0
3 years ago
How much of strontium-90 will be left after 40 hours if you start with 960 grams and the half-life is 10 hours?
zmey [24]

Answer:

5 hours

Explanation:

7 0
2 years ago
If a 250mL beaker weighs 13.473g and the same beaker plus 2.2 mL of water weighs 15.346g. How much does the 2.2 mL of water, alo
Sindrei [870]

Given :

A 250 ml beaker weighs 13.473 g .

The same beaker plus 2.2 ml of water weighs 15.346 g.

To Find :

How much does the 2.2 ml of water, alone, weigh .

Solution :

Now, mass of water is given by :

\text{Mass of water alone = Total mass - Mass of beaker alone}\\\\Mass =15.346-13.473\ g\\\\Mass =1.873\ g

Therefore , mass of 2.2 ml of water alone is 1.873 g .

Hence , this is the required solution .

4 0
3 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
What liquid is got from copper ii and Ammonia gas?​
11111nata11111 [884]

Answer:

deep blue solution of tetramminecopper [Cu(NH3)4]2+ complex ion.

Explanation:

6 0
2 years ago
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