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kondor19780726 [428]
3 years ago
5

25.0 g sample of compound A is mixed with a 40.0 g sample of compound B. a chemical reaction occurs. Once the reaction is comple

te, the final mixture has a mass of 55.0g. What happens to the “missing” mass? How much mass is missing?
Chemistry
1 answer:
artcher [175]3 years ago
4 0

Answer:

Explanation:

Mass of compound A  = 25g

Mass of compound B = 40g

Mass of final mixture = 55g

What happens to the missing mass?

According to the law of conservation of mass, in chemical reaction, matter is transformed from one form to another but cannot be created nor destroyed.

We expect the final mass of the mixture and that of the reacting compounds to be the same but the opposite is the case.

There is a mass loss which typifies most chemical reaction.

The reason for this is that some of the masses must have been lost by the production of gaseous species which are unaccounted for.

The missing mass:

  Total mass expected  = mass of A + mass of B = 25 + 40 = 65g

Missing mass = expected mass - mass of final mixture = 65 - 55 = 10g

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Misha Larkins [42]

For the first part, use the question M=mol/vol (liters)

To do this, you have the given 1.6 M solution

divide the 360g by the molar mass of ethanol (44.07) to get moles

360/44.07=8.16 mol

so

1.6M = 8.16 mol/x vol

volume: 5.1 Liters

8 0
3 years ago
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
What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate
denis23 [38]

1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of  Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

x = 2.76 / 2

x = 1.38

8 0
3 years ago
What is the oxidation state of each element in the species Mn(ClO4)3?
gayaneshka [121]
The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem. For oxygen, the charge is 2-, the total considering its number of atoms is -24. Mn has a charge of +3. TO compute for Mn, we must achieve zero charge overall hence 3+3x-24=0 where x is the Cl charge. Cl charge, x is +7. 
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3 years ago
Read 2 more answers
A chemist needs to make 250 mL of a 2.50 M aqueous solution of ammonium hydroxide from a 6.00 M ammonium hydroxide solution. How
Furkat [3]

Answer:

250 mL (total solution)  = 104 mL (stock solution)  + 146 mL (water)

Explanation:

Data Given

M1 = 6.00 M

M2 = 2.5 M

V1 = 250 mL

V2 = ?

Solution:

As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.

Now

first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution

For this Purpose we use the following formula

                    M1V1=M2V2

Put values from given data in the formula

                   6 x V1 = 2.5 x 250

Rearrange the equation

                   V1 = 2.5 x 250 /6

                    V1 = 104 mL

So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M  aqueous solution of ammonium hydroxide

But we have to prepare 250 mL of the solution.

so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.

in this question you have to tell about the amount of water that is 146 mL

250 mL (total solution)  = 104 mL (stock solution)  + 146 mL (water)

7 0
3 years ago
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